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1 | 7318-7321 | , n) is the probability of the random variable X taking the value xi i e ,
P(X = xi) = pi
© NCERT
not to be republished
PROBABILITY 561
�Note If xi is one of the possible values of a random variable X, the statement
X = xi is true only at some point (s) of the sample space Hence, the probability that
X takes value xi is always nonzero, i |
1 | 7319-7322 | e ,
P(X = xi) = pi
© NCERT
not to be republished
PROBABILITY 561
�Note If xi is one of the possible values of a random variable X, the statement
X = xi is true only at some point (s) of the sample space Hence, the probability that
X takes value xi is always nonzero, i e |
1 | 7320-7323 | ,
P(X = xi) = pi
© NCERT
not to be republished
PROBABILITY 561
�Note If xi is one of the possible values of a random variable X, the statement
X = xi is true only at some point (s) of the sample space Hence, the probability that
X takes value xi is always nonzero, i e P(X = xi) ≠ 0 |
1 | 7321-7324 | Hence, the probability that
X takes value xi is always nonzero, i e P(X = xi) ≠ 0 Also for all possible values of the random variable X, all elements of the sample
space are covered |
1 | 7322-7325 | e P(X = xi) ≠ 0 Also for all possible values of the random variable X, all elements of the sample
space are covered Hence, the sum of all the probabilities in a probability distribution
must be one |
1 | 7323-7326 | P(X = xi) ≠ 0 Also for all possible values of the random variable X, all elements of the sample
space are covered Hence, the sum of all the probabilities in a probability distribution
must be one Example 24 Two cards are drawn successively with replacement from a well-shuffled
deck of 52 cards |
1 | 7324-7327 | Also for all possible values of the random variable X, all elements of the sample
space are covered Hence, the sum of all the probabilities in a probability distribution
must be one Example 24 Two cards are drawn successively with replacement from a well-shuffled
deck of 52 cards Find the probability distribution of the number of aces |
1 | 7325-7328 | Hence, the sum of all the probabilities in a probability distribution
must be one Example 24 Two cards are drawn successively with replacement from a well-shuffled
deck of 52 cards Find the probability distribution of the number of aces Solution The number of aces is a random variable |
1 | 7326-7329 | Example 24 Two cards are drawn successively with replacement from a well-shuffled
deck of 52 cards Find the probability distribution of the number of aces Solution The number of aces is a random variable Let it be denoted by X |
1 | 7327-7330 | Find the probability distribution of the number of aces Solution The number of aces is a random variable Let it be denoted by X Clearly, X
can take the values 0, 1, or 2 |
1 | 7328-7331 | Solution The number of aces is a random variable Let it be denoted by X Clearly, X
can take the values 0, 1, or 2 Now, since the draws are done with replacement, therefore, the two draws form
independent experiments |
1 | 7329-7332 | Let it be denoted by X Clearly, X
can take the values 0, 1, or 2 Now, since the draws are done with replacement, therefore, the two draws form
independent experiments Therefore,
P(X = 0) = P(non-ace and non-ace)
= P(non-ace) × P(non-ace)
= 48
48
144
52
52
169
×
=
P(X = 1) = P(ace and non-ace or non-ace and ace)
= P(ace and non-ace) + P(non-ace and ace)
= P(ace) |
1 | 7330-7333 | Clearly, X
can take the values 0, 1, or 2 Now, since the draws are done with replacement, therefore, the two draws form
independent experiments Therefore,
P(X = 0) = P(non-ace and non-ace)
= P(non-ace) × P(non-ace)
= 48
48
144
52
52
169
×
=
P(X = 1) = P(ace and non-ace or non-ace and ace)
= P(ace and non-ace) + P(non-ace and ace)
= P(ace) P(non-ace) + P (non-ace) |
1 | 7331-7334 | Now, since the draws are done with replacement, therefore, the two draws form
independent experiments Therefore,
P(X = 0) = P(non-ace and non-ace)
= P(non-ace) × P(non-ace)
= 48
48
144
52
52
169
×
=
P(X = 1) = P(ace and non-ace or non-ace and ace)
= P(ace and non-ace) + P(non-ace and ace)
= P(ace) P(non-ace) + P (non-ace) P(ace)
= 4
48
48
4
24
52
52
52
52
169
×
+
×
=
and
P(X = 2) = P (ace and ace)
= 4
4
1
52
52
169
Thus, the required probability distribution is
X
0
1
2
P(X)
144
169
24
169
1
169
Example 25 Find the probability distribution of number of doublets in three throws of
a pair of dice |
1 | 7332-7335 | Therefore,
P(X = 0) = P(non-ace and non-ace)
= P(non-ace) × P(non-ace)
= 48
48
144
52
52
169
×
=
P(X = 1) = P(ace and non-ace or non-ace and ace)
= P(ace and non-ace) + P(non-ace and ace)
= P(ace) P(non-ace) + P (non-ace) P(ace)
= 4
48
48
4
24
52
52
52
52
169
×
+
×
=
and
P(X = 2) = P (ace and ace)
= 4
4
1
52
52
169
Thus, the required probability distribution is
X
0
1
2
P(X)
144
169
24
169
1
169
Example 25 Find the probability distribution of number of doublets in three throws of
a pair of dice © NCERT
not to be republished
562
MATHEMATICS
Solution Let X denote the number of doublets |
1 | 7333-7336 | P(non-ace) + P (non-ace) P(ace)
= 4
48
48
4
24
52
52
52
52
169
×
+
×
=
and
P(X = 2) = P (ace and ace)
= 4
4
1
52
52
169
Thus, the required probability distribution is
X
0
1
2
P(X)
144
169
24
169
1
169
Example 25 Find the probability distribution of number of doublets in three throws of
a pair of dice © NCERT
not to be republished
562
MATHEMATICS
Solution Let X denote the number of doublets Possible doublets are
(1,1) , (2,2), (3,3), (4,4), (5,5), (6,6)
Clearly, X can take the value 0, 1, 2, or 3 |
1 | 7334-7337 | P(ace)
= 4
48
48
4
24
52
52
52
52
169
×
+
×
=
and
P(X = 2) = P (ace and ace)
= 4
4
1
52
52
169
Thus, the required probability distribution is
X
0
1
2
P(X)
144
169
24
169
1
169
Example 25 Find the probability distribution of number of doublets in three throws of
a pair of dice © NCERT
not to be republished
562
MATHEMATICS
Solution Let X denote the number of doublets Possible doublets are
(1,1) , (2,2), (3,3), (4,4), (5,5), (6,6)
Clearly, X can take the value 0, 1, 2, or 3 Probability of getting a doublet
6
1
36
6
Probability of not getting a doublet
1
5
1
6
6
Now
P(X = 0) = P (no doublet) = 5
5
5
125
6
6
6
216
P(X = 1) = P (one doublet and two non-doublets)
= 1
5
5
5
1
5
5
5
1
6
6
6
6
6
6
6
6
6
=
2
2
1
5
75
3 6
216
6
P(X = 2) = P (two doublets and one non-doublet)
=
2
1
1
5
1
5
1
5
1
1
1
5
15
3
6
6
6
6
6
6
6
6
6
6
216
6
and
P(X = 3) = P (three doublets)
= 1
1
1
1
6
6
6
216
Thus, the required probability distribution is
X
0
1
2
3
P(X)
125
216
75
216
15
216
1
216
Verification Sum of the probabilities
1
n
i
i
p
=∑
= 125
75
15
1
216
216
216
216
= 125
75
15
1
216
1
216
216
© NCERT
not to be republished
PROBABILITY 563
Example 26 Let X denote the number of hours you study during a randomly selected
school day |
1 | 7335-7338 | © NCERT
not to be republished
562
MATHEMATICS
Solution Let X denote the number of doublets Possible doublets are
(1,1) , (2,2), (3,3), (4,4), (5,5), (6,6)
Clearly, X can take the value 0, 1, 2, or 3 Probability of getting a doublet
6
1
36
6
Probability of not getting a doublet
1
5
1
6
6
Now
P(X = 0) = P (no doublet) = 5
5
5
125
6
6
6
216
P(X = 1) = P (one doublet and two non-doublets)
= 1
5
5
5
1
5
5
5
1
6
6
6
6
6
6
6
6
6
=
2
2
1
5
75
3 6
216
6
P(X = 2) = P (two doublets and one non-doublet)
=
2
1
1
5
1
5
1
5
1
1
1
5
15
3
6
6
6
6
6
6
6
6
6
6
216
6
and
P(X = 3) = P (three doublets)
= 1
1
1
1
6
6
6
216
Thus, the required probability distribution is
X
0
1
2
3
P(X)
125
216
75
216
15
216
1
216
Verification Sum of the probabilities
1
n
i
i
p
=∑
= 125
75
15
1
216
216
216
216
= 125
75
15
1
216
1
216
216
© NCERT
not to be republished
PROBABILITY 563
Example 26 Let X denote the number of hours you study during a randomly selected
school day The probability that X can take the values x, has the following form, where
k is some unknown constant |
1 | 7336-7339 | Possible doublets are
(1,1) , (2,2), (3,3), (4,4), (5,5), (6,6)
Clearly, X can take the value 0, 1, 2, or 3 Probability of getting a doublet
6
1
36
6
Probability of not getting a doublet
1
5
1
6
6
Now
P(X = 0) = P (no doublet) = 5
5
5
125
6
6
6
216
P(X = 1) = P (one doublet and two non-doublets)
= 1
5
5
5
1
5
5
5
1
6
6
6
6
6
6
6
6
6
=
2
2
1
5
75
3 6
216
6
P(X = 2) = P (two doublets and one non-doublet)
=
2
1
1
5
1
5
1
5
1
1
1
5
15
3
6
6
6
6
6
6
6
6
6
6
216
6
and
P(X = 3) = P (three doublets)
= 1
1
1
1
6
6
6
216
Thus, the required probability distribution is
X
0
1
2
3
P(X)
125
216
75
216
15
216
1
216
Verification Sum of the probabilities
1
n
i
i
p
=∑
= 125
75
15
1
216
216
216
216
= 125
75
15
1
216
1
216
216
© NCERT
not to be republished
PROBABILITY 563
Example 26 Let X denote the number of hours you study during a randomly selected
school day The probability that X can take the values x, has the following form, where
k is some unknown constant P(X = x) =
0 |
1 | 7337-7340 | Probability of getting a doublet
6
1
36
6
Probability of not getting a doublet
1
5
1
6
6
Now
P(X = 0) = P (no doublet) = 5
5
5
125
6
6
6
216
P(X = 1) = P (one doublet and two non-doublets)
= 1
5
5
5
1
5
5
5
1
6
6
6
6
6
6
6
6
6
=
2
2
1
5
75
3 6
216
6
P(X = 2) = P (two doublets and one non-doublet)
=
2
1
1
5
1
5
1
5
1
1
1
5
15
3
6
6
6
6
6
6
6
6
6
6
216
6
and
P(X = 3) = P (three doublets)
= 1
1
1
1
6
6
6
216
Thus, the required probability distribution is
X
0
1
2
3
P(X)
125
216
75
216
15
216
1
216
Verification Sum of the probabilities
1
n
i
i
p
=∑
= 125
75
15
1
216
216
216
216
= 125
75
15
1
216
1
216
216
© NCERT
not to be republished
PROBABILITY 563
Example 26 Let X denote the number of hours you study during a randomly selected
school day The probability that X can take the values x, has the following form, where
k is some unknown constant P(X = x) =
0 1, if
0
, if
1or2
(5
), if
3or4
0, otherwise
=
⎪⎧
=
⎪⎨
−
=
⎪
⎪⎩
x
kx
x
k
x
x
(a)
Find the value of k |
1 | 7338-7341 | The probability that X can take the values x, has the following form, where
k is some unknown constant P(X = x) =
0 1, if
0
, if
1or2
(5
), if
3or4
0, otherwise
=
⎪⎧
=
⎪⎨
−
=
⎪
⎪⎩
x
kx
x
k
x
x
(a)
Find the value of k (b)
What is the probability that you study at least two hours |
1 | 7339-7342 | P(X = x) =
0 1, if
0
, if
1or2
(5
), if
3or4
0, otherwise
=
⎪⎧
=
⎪⎨
−
=
⎪
⎪⎩
x
kx
x
k
x
x
(a)
Find the value of k (b)
What is the probability that you study at least two hours Exactly two hours |
1 | 7340-7343 | 1, if
0
, if
1or2
(5
), if
3or4
0, otherwise
=
⎪⎧
=
⎪⎨
−
=
⎪
⎪⎩
x
kx
x
k
x
x
(a)
Find the value of k (b)
What is the probability that you study at least two hours Exactly two hours At
most two hours |
1 | 7341-7344 | (b)
What is the probability that you study at least two hours Exactly two hours At
most two hours Solution The probability distribution of X is
X
0
1
2
3
4
P(X)
0 |
1 | 7342-7345 | Exactly two hours At
most two hours Solution The probability distribution of X is
X
0
1
2
3
4
P(X)
0 1
k
2k
2k
k
(a)
We know that
1
n
i
i
p
=∑
= 1
Therefore
0 |
1 | 7343-7346 | At
most two hours Solution The probability distribution of X is
X
0
1
2
3
4
P(X)
0 1
k
2k
2k
k
(a)
We know that
1
n
i
i
p
=∑
= 1
Therefore
0 1 + k + 2k + 2k + k = 1
i |
1 | 7344-7347 | Solution The probability distribution of X is
X
0
1
2
3
4
P(X)
0 1
k
2k
2k
k
(a)
We know that
1
n
i
i
p
=∑
= 1
Therefore
0 1 + k + 2k + 2k + k = 1
i e |
1 | 7345-7348 | 1
k
2k
2k
k
(a)
We know that
1
n
i
i
p
=∑
= 1
Therefore
0 1 + k + 2k + 2k + k = 1
i e k = 0 |
1 | 7346-7349 | 1 + k + 2k + 2k + k = 1
i e k = 0 15
(b)
P(you study at least two hours)
= P(X ≥ 2)
= P(X = 2) + P (X = 3) + P (X = 4)
= 2k + 2k + k = 5k = 5 × 0 |
1 | 7347-7350 | e k = 0 15
(b)
P(you study at least two hours)
= P(X ≥ 2)
= P(X = 2) + P (X = 3) + P (X = 4)
= 2k + 2k + k = 5k = 5 × 0 15 = 0 |
1 | 7348-7351 | k = 0 15
(b)
P(you study at least two hours)
= P(X ≥ 2)
= P(X = 2) + P (X = 3) + P (X = 4)
= 2k + 2k + k = 5k = 5 × 0 15 = 0 75
P(you study exactly two hours)
= P(X = 2)
= 2k = 2 × 0 |
1 | 7349-7352 | 15
(b)
P(you study at least two hours)
= P(X ≥ 2)
= P(X = 2) + P (X = 3) + P (X = 4)
= 2k + 2k + k = 5k = 5 × 0 15 = 0 75
P(you study exactly two hours)
= P(X = 2)
= 2k = 2 × 0 15 = 0 |
1 | 7350-7353 | 15 = 0 75
P(you study exactly two hours)
= P(X = 2)
= 2k = 2 × 0 15 = 0 3
P(you study at most two hours)
= P(X ≤ 2)
= P (X = 0) + P(X = 1) + P(X = 2)
= 0 |
1 | 7351-7354 | 75
P(you study exactly two hours)
= P(X = 2)
= 2k = 2 × 0 15 = 0 3
P(you study at most two hours)
= P(X ≤ 2)
= P (X = 0) + P(X = 1) + P(X = 2)
= 0 1 + k + 2k = 0 |
1 | 7352-7355 | 15 = 0 3
P(you study at most two hours)
= P(X ≤ 2)
= P (X = 0) + P(X = 1) + P(X = 2)
= 0 1 + k + 2k = 0 1 + 3k = 0 |
1 | 7353-7356 | 3
P(you study at most two hours)
= P(X ≤ 2)
= P (X = 0) + P(X = 1) + P(X = 2)
= 0 1 + k + 2k = 0 1 + 3k = 0 1 + 3 × 0 |
1 | 7354-7357 | 1 + k + 2k = 0 1 + 3k = 0 1 + 3 × 0 15
= 0 |
1 | 7355-7358 | 1 + 3k = 0 1 + 3 × 0 15
= 0 55
13 |
1 | 7356-7359 | 1 + 3 × 0 15
= 0 55
13 6 |
1 | 7357-7360 | 15
= 0 55
13 6 2 Mean of a random variable
In many problems, it is desirable to describe some feature of the random variable by
means of a single number that can be computed from its probability distribution |
1 | 7358-7361 | 55
13 6 2 Mean of a random variable
In many problems, it is desirable to describe some feature of the random variable by
means of a single number that can be computed from its probability distribution Few
such numbers are mean, median and mode |
1 | 7359-7362 | 6 2 Mean of a random variable
In many problems, it is desirable to describe some feature of the random variable by
means of a single number that can be computed from its probability distribution Few
such numbers are mean, median and mode In this section, we shall discuss mean only |
1 | 7360-7363 | 2 Mean of a random variable
In many problems, it is desirable to describe some feature of the random variable by
means of a single number that can be computed from its probability distribution Few
such numbers are mean, median and mode In this section, we shall discuss mean only Mean is a measure of location or central tendency in the sense that it roughly locates a
middle or average value of the random variable |
1 | 7361-7364 | Few
such numbers are mean, median and mode In this section, we shall discuss mean only Mean is a measure of location or central tendency in the sense that it roughly locates a
middle or average value of the random variable © NCERT
not to be republished
564
MATHEMATICS
Definition 6 Let X be a random variable whose possible values x1, x2, x3, |
1 | 7362-7365 | In this section, we shall discuss mean only Mean is a measure of location or central tendency in the sense that it roughly locates a
middle or average value of the random variable © NCERT
not to be republished
564
MATHEMATICS
Definition 6 Let X be a random variable whose possible values x1, x2, x3, , xn occur
with probabilities p1, p2, p3, |
1 | 7363-7366 | Mean is a measure of location or central tendency in the sense that it roughly locates a
middle or average value of the random variable © NCERT
not to be republished
564
MATHEMATICS
Definition 6 Let X be a random variable whose possible values x1, x2, x3, , xn occur
with probabilities p1, p2, p3, , pn, respectively |
1 | 7364-7367 | © NCERT
not to be republished
564
MATHEMATICS
Definition 6 Let X be a random variable whose possible values x1, x2, x3, , xn occur
with probabilities p1, p2, p3, , pn, respectively The mean of X, denoted by μ, is the
number
1
n
i
i
i
x p
=∑
i |
1 | 7365-7368 | , xn occur
with probabilities p1, p2, p3, , pn, respectively The mean of X, denoted by μ, is the
number
1
n
i
i
i
x p
=∑
i e |
1 | 7366-7369 | , pn, respectively The mean of X, denoted by μ, is the
number
1
n
i
i
i
x p
=∑
i e the mean of X is the weighted average of the possible values of X,
each value being weighted by its probability with which it occurs |
1 | 7367-7370 | The mean of X, denoted by μ, is the
number
1
n
i
i
i
x p
=∑
i e the mean of X is the weighted average of the possible values of X,
each value being weighted by its probability with which it occurs The mean of a random variable X is also called the expectation of X, denoted by
E(X) |
1 | 7368-7371 | e the mean of X is the weighted average of the possible values of X,
each value being weighted by its probability with which it occurs The mean of a random variable X is also called the expectation of X, denoted by
E(X) Thus,
E (X) = μ =
1
n
i
i
i
x p
= x1 p1+ x2 p2 + |
1 | 7369-7372 | the mean of X is the weighted average of the possible values of X,
each value being weighted by its probability with which it occurs The mean of a random variable X is also called the expectation of X, denoted by
E(X) Thus,
E (X) = μ =
1
n
i
i
i
x p
= x1 p1+ x2 p2 + + xn pn |
1 | 7370-7373 | The mean of a random variable X is also called the expectation of X, denoted by
E(X) Thus,
E (X) = μ =
1
n
i
i
i
x p
= x1 p1+ x2 p2 + + xn pn In other words, the mean or expectation of a random variable X is the sum of the
products of all possible values of X by their respective probabilities |
1 | 7371-7374 | Thus,
E (X) = μ =
1
n
i
i
i
x p
= x1 p1+ x2 p2 + + xn pn In other words, the mean or expectation of a random variable X is the sum of the
products of all possible values of X by their respective probabilities Example 27 Let a pair of dice be thrown and the random variable X be the sum of the
numbers that appear on the two dice |
1 | 7372-7375 | + xn pn In other words, the mean or expectation of a random variable X is the sum of the
products of all possible values of X by their respective probabilities Example 27 Let a pair of dice be thrown and the random variable X be the sum of the
numbers that appear on the two dice Find the mean or expectation of X |
1 | 7373-7376 | In other words, the mean or expectation of a random variable X is the sum of the
products of all possible values of X by their respective probabilities Example 27 Let a pair of dice be thrown and the random variable X be the sum of the
numbers that appear on the two dice Find the mean or expectation of X Solution The sample space of the experiment consists of 36 elementary events in the
form of ordered pairs (xi, yi), where xi = 1, 2, 3, 4, 5, 6 and yi = 1, 2, 3, 4, 5, 6 |
1 | 7374-7377 | Example 27 Let a pair of dice be thrown and the random variable X be the sum of the
numbers that appear on the two dice Find the mean or expectation of X Solution The sample space of the experiment consists of 36 elementary events in the
form of ordered pairs (xi, yi), where xi = 1, 2, 3, 4, 5, 6 and yi = 1, 2, 3, 4, 5, 6 The random variable X i |
1 | 7375-7378 | Find the mean or expectation of X Solution The sample space of the experiment consists of 36 elementary events in the
form of ordered pairs (xi, yi), where xi = 1, 2, 3, 4, 5, 6 and yi = 1, 2, 3, 4, 5, 6 The random variable X i e |
1 | 7376-7379 | Solution The sample space of the experiment consists of 36 elementary events in the
form of ordered pairs (xi, yi), where xi = 1, 2, 3, 4, 5, 6 and yi = 1, 2, 3, 4, 5, 6 The random variable X i e the sum of the numbers on the two dice takes the
values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12 |
1 | 7377-7380 | The random variable X i e the sum of the numbers on the two dice takes the
values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12 Now
P(X = 2) = P({(1,1)})
361
P(X = 3) = P({(1,2), (2,1)})
362
P(X = 4) = P({(1,3), (2,2), (3,1)})
363
P(X = 5) = P({(1,4), (2,3), (3,2), (4,1)})
364
P(X = 6) = P({(1,5), (2,4), (3,3), (4,2), (5,1)})
365
P(X = 7) = P({(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)})
366
P(X = 8) = P({(2,6), (3,5), (4,4), (5,3), (6,2)})
365
© NCERT
not to be republished
PROBABILITY 565
P(X = 9) = P({(3,6), (4,5), (5,4), (6,3)})
364
P(X = 10) = P({(4,6), (5,5), (6,4)})
363
P(X = 11) = P({(5,6), (6,5)})
362
P(X = 12) = P({(6,6)})
361
The probability distribution of X is
X or xi
2
3
4
5
6
7
8
9
10
11
12
P(X) or pi
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Therefore,
μ = E(X) =
1
1
2
3
4
2
3
4
5
36
36
36
36
n
i
i
i
x p
=
= ×
+ ×
+ ×
+ ×
∑
5
6
5
6
7
8
36
36
36
4
3
2
1
9
10
11
12
36
36
36
36
= 2
6
12
20
30
42
40
36
30
22
12
36
= 7
Thus, the mean of the sum of the numbers that appear on throwing two fair dice is 7 |
1 | 7378-7381 | e the sum of the numbers on the two dice takes the
values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12 Now
P(X = 2) = P({(1,1)})
361
P(X = 3) = P({(1,2), (2,1)})
362
P(X = 4) = P({(1,3), (2,2), (3,1)})
363
P(X = 5) = P({(1,4), (2,3), (3,2), (4,1)})
364
P(X = 6) = P({(1,5), (2,4), (3,3), (4,2), (5,1)})
365
P(X = 7) = P({(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)})
366
P(X = 8) = P({(2,6), (3,5), (4,4), (5,3), (6,2)})
365
© NCERT
not to be republished
PROBABILITY 565
P(X = 9) = P({(3,6), (4,5), (5,4), (6,3)})
364
P(X = 10) = P({(4,6), (5,5), (6,4)})
363
P(X = 11) = P({(5,6), (6,5)})
362
P(X = 12) = P({(6,6)})
361
The probability distribution of X is
X or xi
2
3
4
5
6
7
8
9
10
11
12
P(X) or pi
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Therefore,
μ = E(X) =
1
1
2
3
4
2
3
4
5
36
36
36
36
n
i
i
i
x p
=
= ×
+ ×
+ ×
+ ×
∑
5
6
5
6
7
8
36
36
36
4
3
2
1
9
10
11
12
36
36
36
36
= 2
6
12
20
30
42
40
36
30
22
12
36
= 7
Thus, the mean of the sum of the numbers that appear on throwing two fair dice is 7 13 |
1 | 7379-7382 | the sum of the numbers on the two dice takes the
values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12 Now
P(X = 2) = P({(1,1)})
361
P(X = 3) = P({(1,2), (2,1)})
362
P(X = 4) = P({(1,3), (2,2), (3,1)})
363
P(X = 5) = P({(1,4), (2,3), (3,2), (4,1)})
364
P(X = 6) = P({(1,5), (2,4), (3,3), (4,2), (5,1)})
365
P(X = 7) = P({(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)})
366
P(X = 8) = P({(2,6), (3,5), (4,4), (5,3), (6,2)})
365
© NCERT
not to be republished
PROBABILITY 565
P(X = 9) = P({(3,6), (4,5), (5,4), (6,3)})
364
P(X = 10) = P({(4,6), (5,5), (6,4)})
363
P(X = 11) = P({(5,6), (6,5)})
362
P(X = 12) = P({(6,6)})
361
The probability distribution of X is
X or xi
2
3
4
5
6
7
8
9
10
11
12
P(X) or pi
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Therefore,
μ = E(X) =
1
1
2
3
4
2
3
4
5
36
36
36
36
n
i
i
i
x p
=
= ×
+ ×
+ ×
+ ×
∑
5
6
5
6
7
8
36
36
36
4
3
2
1
9
10
11
12
36
36
36
36
= 2
6
12
20
30
42
40
36
30
22
12
36
= 7
Thus, the mean of the sum of the numbers that appear on throwing two fair dice is 7 13 6 |
1 | 7380-7383 | Now
P(X = 2) = P({(1,1)})
361
P(X = 3) = P({(1,2), (2,1)})
362
P(X = 4) = P({(1,3), (2,2), (3,1)})
363
P(X = 5) = P({(1,4), (2,3), (3,2), (4,1)})
364
P(X = 6) = P({(1,5), (2,4), (3,3), (4,2), (5,1)})
365
P(X = 7) = P({(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)})
366
P(X = 8) = P({(2,6), (3,5), (4,4), (5,3), (6,2)})
365
© NCERT
not to be republished
PROBABILITY 565
P(X = 9) = P({(3,6), (4,5), (5,4), (6,3)})
364
P(X = 10) = P({(4,6), (5,5), (6,4)})
363
P(X = 11) = P({(5,6), (6,5)})
362
P(X = 12) = P({(6,6)})
361
The probability distribution of X is
X or xi
2
3
4
5
6
7
8
9
10
11
12
P(X) or pi
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Therefore,
μ = E(X) =
1
1
2
3
4
2
3
4
5
36
36
36
36
n
i
i
i
x p
=
= ×
+ ×
+ ×
+ ×
∑
5
6
5
6
7
8
36
36
36
4
3
2
1
9
10
11
12
36
36
36
36
= 2
6
12
20
30
42
40
36
30
22
12
36
= 7
Thus, the mean of the sum of the numbers that appear on throwing two fair dice is 7 13 6 3 Variance of a random variable
The mean of a random variable does not give us information about the variability in the
values of the random variable |
1 | 7381-7384 | 13 6 3 Variance of a random variable
The mean of a random variable does not give us information about the variability in the
values of the random variable In fact, if the variance is small, then the values of the
random variable are close to the mean |
1 | 7382-7385 | 6 3 Variance of a random variable
The mean of a random variable does not give us information about the variability in the
values of the random variable In fact, if the variance is small, then the values of the
random variable are close to the mean Also random variables with different probability
distributions can have equal means, as shown in the following distributions of X and Y |
1 | 7383-7386 | 3 Variance of a random variable
The mean of a random variable does not give us information about the variability in the
values of the random variable In fact, if the variance is small, then the values of the
random variable are close to the mean Also random variables with different probability
distributions can have equal means, as shown in the following distributions of X and Y X
1
2
3
4
P(X)
1
8
2
8
3
8
2
8
© NCERT
not to be republished
566
MATHEMATICS
Y
–1
0
4
5
6
P(Y)
1
8
2
8
3
8
1
8
1
8
Clearly
E(X) =
1
2
3
2
22
1
2
3
4
2 |
1 | 7384-7387 | In fact, if the variance is small, then the values of the
random variable are close to the mean Also random variables with different probability
distributions can have equal means, as shown in the following distributions of X and Y X
1
2
3
4
P(X)
1
8
2
8
3
8
2
8
© NCERT
not to be republished
566
MATHEMATICS
Y
–1
0
4
5
6
P(Y)
1
8
2
8
3
8
1
8
1
8
Clearly
E(X) =
1
2
3
2
22
1
2
3
4
2 75
8
8
8
8
8
× + ×
+ × + ×
=
=
and
E(Y) =
1
2
3
1
1
22
1
0
4
5
6
2 |
1 | 7385-7388 | Also random variables with different probability
distributions can have equal means, as shown in the following distributions of X and Y X
1
2
3
4
P(X)
1
8
2
8
3
8
2
8
© NCERT
not to be republished
566
MATHEMATICS
Y
–1
0
4
5
6
P(Y)
1
8
2
8
3
8
1
8
1
8
Clearly
E(X) =
1
2
3
2
22
1
2
3
4
2 75
8
8
8
8
8
× + ×
+ × + ×
=
=
and
E(Y) =
1
2
3
1
1
22
1
0
4
5
6
2 75
8
8
8
8
8
8
− × + ×
+ × + × =
× =
=
The variables X and Y are different, however their means are same |
1 | 7386-7389 | X
1
2
3
4
P(X)
1
8
2
8
3
8
2
8
© NCERT
not to be republished
566
MATHEMATICS
Y
–1
0
4
5
6
P(Y)
1
8
2
8
3
8
1
8
1
8
Clearly
E(X) =
1
2
3
2
22
1
2
3
4
2 75
8
8
8
8
8
× + ×
+ × + ×
=
=
and
E(Y) =
1
2
3
1
1
22
1
0
4
5
6
2 75
8
8
8
8
8
8
− × + ×
+ × + × =
× =
=
The variables X and Y are different, however their means are same It is also
easily observable from the diagramatic representation of these distributions (Fig 13 |
1 | 7387-7390 | 75
8
8
8
8
8
× + ×
+ × + ×
=
=
and
E(Y) =
1
2
3
1
1
22
1
0
4
5
6
2 75
8
8
8
8
8
8
− × + ×
+ × + × =
× =
=
The variables X and Y are different, however their means are same It is also
easily observable from the diagramatic representation of these distributions (Fig 13 5) |
1 | 7388-7391 | 75
8
8
8
8
8
8
− × + ×
+ × + × =
× =
=
The variables X and Y are different, however their means are same It is also
easily observable from the diagramatic representation of these distributions (Fig 13 5) Fig 13 |
1 | 7389-7392 | It is also
easily observable from the diagramatic representation of these distributions (Fig 13 5) Fig 13 5
To distinguish X from Y, we require a measure of the extent to which the values of
the random variables spread out |
1 | 7390-7393 | 5) Fig 13 5
To distinguish X from Y, we require a measure of the extent to which the values of
the random variables spread out In Statistics, we have studied that the variance is a
measure of the spread or scatter in data |
1 | 7391-7394 | Fig 13 5
To distinguish X from Y, we require a measure of the extent to which the values of
the random variables spread out In Statistics, we have studied that the variance is a
measure of the spread or scatter in data Likewise, the variability or spread in the
values of a random variable may be measured by variance |
1 | 7392-7395 | 5
To distinguish X from Y, we require a measure of the extent to which the values of
the random variables spread out In Statistics, we have studied that the variance is a
measure of the spread or scatter in data Likewise, the variability or spread in the
values of a random variable may be measured by variance Definition 7 Let X be a random variable whose possible values x1, x2, |
1 | 7393-7396 | In Statistics, we have studied that the variance is a
measure of the spread or scatter in data Likewise, the variability or spread in the
values of a random variable may be measured by variance Definition 7 Let X be a random variable whose possible values x1, x2, ,xn occur with
probabilities p(x1), p(x2), |
1 | 7394-7397 | Likewise, the variability or spread in the
values of a random variable may be measured by variance Definition 7 Let X be a random variable whose possible values x1, x2, ,xn occur with
probabilities p(x1), p(x2), , p(xn) respectively |
1 | 7395-7398 | Definition 7 Let X be a random variable whose possible values x1, x2, ,xn occur with
probabilities p(x1), p(x2), , p(xn) respectively Let μ = E (X) be the mean of X |
1 | 7396-7399 | ,xn occur with
probabilities p(x1), p(x2), , p(xn) respectively Let μ = E (X) be the mean of X The variance of X, denoted by Var (X) or
2
x
is
defined as
2
σx
=
2
1
Var(X)=
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
or equivalently
2
x
= E(X – μ)2
O
1 8
2 8
3 8
P(Y)
O
1 8
2 8
3 8
P(X)
1
2
3
4
1
2
3
4
–1
5
6
(i)
(ii)
© NCERT
not to be republished
PROBABILITY 567
The non-negative number
σx =
2
1
Var(X) =
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
is called the standard deviation of the random variable X |
1 | 7397-7400 | , p(xn) respectively Let μ = E (X) be the mean of X The variance of X, denoted by Var (X) or
2
x
is
defined as
2
σx
=
2
1
Var(X)=
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
or equivalently
2
x
= E(X – μ)2
O
1 8
2 8
3 8
P(Y)
O
1 8
2 8
3 8
P(X)
1
2
3
4
1
2
3
4
–1
5
6
(i)
(ii)
© NCERT
not to be republished
PROBABILITY 567
The non-negative number
σx =
2
1
Var(X) =
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
is called the standard deviation of the random variable X Another formula to find the variance of a random variable |
1 | 7398-7401 | Let μ = E (X) be the mean of X The variance of X, denoted by Var (X) or
2
x
is
defined as
2
σx
=
2
1
Var(X)=
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
or equivalently
2
x
= E(X – μ)2
O
1 8
2 8
3 8
P(Y)
O
1 8
2 8
3 8
P(X)
1
2
3
4
1
2
3
4
–1
5
6
(i)
(ii)
© NCERT
not to be republished
PROBABILITY 567
The non-negative number
σx =
2
1
Var(X) =
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
is called the standard deviation of the random variable X Another formula to find the variance of a random variable We know that,
Var (X) =
2
1
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
=
2
2
1
(
μ
2μ
) (
)
n
i
i
i
i
x
x
p x
=
2
2
1
1
1
(
)
μ
(
)
2μ
(
)
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
=
+
−
∑
∑
∑
=
2
2
1
1
1
(
) μ
(
)
2μ
(
)
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
=
+
−
∑
∑
∑
=
2
2
2
1
=1
1
(
) μ
2μ
since
( )=1andμ=
( )
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
⎡
⎤
+
−
⎢
⎥
⎣
⎦
∑
∑
∑
=
2
2
1
(
) μ
n
i
i
i
x
p x
=
−
∑
or
Var (X) =
2
2
1
1
(
)
(
)
n
n
i
i
i
i
i
i
x
p x
x p x
=
=
⎛
⎞
−⎜
⎟
⎝
⎠
∑
∑
or
Var (X) = E(X2) – [E(X)]2, where E(X2) =
2
1
(
)
n
i
i
i
x
p x
=∑
Example 28 Find the variance of the number obtained on a throw of an unbiased die |
1 | 7399-7402 | The variance of X, denoted by Var (X) or
2
x
is
defined as
2
σx
=
2
1
Var(X)=
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
or equivalently
2
x
= E(X – μ)2
O
1 8
2 8
3 8
P(Y)
O
1 8
2 8
3 8
P(X)
1
2
3
4
1
2
3
4
–1
5
6
(i)
(ii)
© NCERT
not to be republished
PROBABILITY 567
The non-negative number
σx =
2
1
Var(X) =
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
is called the standard deviation of the random variable X Another formula to find the variance of a random variable We know that,
Var (X) =
2
1
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
=
2
2
1
(
μ
2μ
) (
)
n
i
i
i
i
x
x
p x
=
2
2
1
1
1
(
)
μ
(
)
2μ
(
)
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
=
+
−
∑
∑
∑
=
2
2
1
1
1
(
) μ
(
)
2μ
(
)
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
=
+
−
∑
∑
∑
=
2
2
2
1
=1
1
(
) μ
2μ
since
( )=1andμ=
( )
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
⎡
⎤
+
−
⎢
⎥
⎣
⎦
∑
∑
∑
=
2
2
1
(
) μ
n
i
i
i
x
p x
=
−
∑
or
Var (X) =
2
2
1
1
(
)
(
)
n
n
i
i
i
i
i
i
x
p x
x p x
=
=
⎛
⎞
−⎜
⎟
⎝
⎠
∑
∑
or
Var (X) = E(X2) – [E(X)]2, where E(X2) =
2
1
(
)
n
i
i
i
x
p x
=∑
Example 28 Find the variance of the number obtained on a throw of an unbiased die Solution The sample space of the experiment is S = {1, 2, 3, 4, 5, 6} |
1 | 7400-7403 | Another formula to find the variance of a random variable We know that,
Var (X) =
2
1
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
=
2
2
1
(
μ
2μ
) (
)
n
i
i
i
i
x
x
p x
=
2
2
1
1
1
(
)
μ
(
)
2μ
(
)
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
=
+
−
∑
∑
∑
=
2
2
1
1
1
(
) μ
(
)
2μ
(
)
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
=
+
−
∑
∑
∑
=
2
2
2
1
=1
1
(
) μ
2μ
since
( )=1andμ=
( )
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
⎡
⎤
+
−
⎢
⎥
⎣
⎦
∑
∑
∑
=
2
2
1
(
) μ
n
i
i
i
x
p x
=
−
∑
or
Var (X) =
2
2
1
1
(
)
(
)
n
n
i
i
i
i
i
i
x
p x
x p x
=
=
⎛
⎞
−⎜
⎟
⎝
⎠
∑
∑
or
Var (X) = E(X2) – [E(X)]2, where E(X2) =
2
1
(
)
n
i
i
i
x
p x
=∑
Example 28 Find the variance of the number obtained on a throw of an unbiased die Solution The sample space of the experiment is S = {1, 2, 3, 4, 5, 6} Let X denote the number obtained on the throw |
1 | 7401-7404 | We know that,
Var (X) =
2
1
(
μ)
(
)
n
i
i
i
x
p x
=
−
∑
=
2
2
1
(
μ
2μ
) (
)
n
i
i
i
i
x
x
p x
=
2
2
1
1
1
(
)
μ
(
)
2μ
(
)
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
=
+
−
∑
∑
∑
=
2
2
1
1
1
(
) μ
(
)
2μ
(
)
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
=
+
−
∑
∑
∑
=
2
2
2
1
=1
1
(
) μ
2μ
since
( )=1andμ=
( )
n
n
n
i
i
i
i
i
i
i
i
x
p x
p x
x p x
=
=
⎡
⎤
+
−
⎢
⎥
⎣
⎦
∑
∑
∑
=
2
2
1
(
) μ
n
i
i
i
x
p x
=
−
∑
or
Var (X) =
2
2
1
1
(
)
(
)
n
n
i
i
i
i
i
i
x
p x
x p x
=
=
⎛
⎞
−⎜
⎟
⎝
⎠
∑
∑
or
Var (X) = E(X2) – [E(X)]2, where E(X2) =
2
1
(
)
n
i
i
i
x
p x
=∑
Example 28 Find the variance of the number obtained on a throw of an unbiased die Solution The sample space of the experiment is S = {1, 2, 3, 4, 5, 6} Let X denote the number obtained on the throw Then X is a random variable
which can take values 1, 2, 3, 4, 5, or 6 |
1 | 7402-7405 | Solution The sample space of the experiment is S = {1, 2, 3, 4, 5, 6} Let X denote the number obtained on the throw Then X is a random variable
which can take values 1, 2, 3, 4, 5, or 6 © NCERT
not to be republished
568
MATHEMATICS
Also
P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1
6
Therefore, the Probability distribution of X is
X
1
2
3
4
5
6
P(X)
1
6
1
6
1
6
1
6
1
6
1
6
Now
E(X) =
1
(
)
n
i
i
i
x p x
=
1
1
1
1
1
1
21
1
2
3
4
5
6
6
6
6
6
6
6
6
×
+ ×
+ ×
+ ×
+ ×
+ ×
=
Also
E(X2) =
2
2
2
2
2
2
1
1
1
1
1
1
91
1
2
3
4
5
6
6
6
6
6
6
6
6
×
+
×
+
×
+
×
+
×
+
×
=
Thus,
Var (X) = E (X2) – (E(X))2
=
2
91
21
91
441
6
6
6
36
⎛
⎞
−
=
−
⎜
⎟
⎝
⎠
35
12
Example 29 Two cards are drawn simultaneously (or successively without replacement)
from a well shuffled pack of 52 cards |
1 | 7403-7406 | Let X denote the number obtained on the throw Then X is a random variable
which can take values 1, 2, 3, 4, 5, or 6 © NCERT
not to be republished
568
MATHEMATICS
Also
P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1
6
Therefore, the Probability distribution of X is
X
1
2
3
4
5
6
P(X)
1
6
1
6
1
6
1
6
1
6
1
6
Now
E(X) =
1
(
)
n
i
i
i
x p x
=
1
1
1
1
1
1
21
1
2
3
4
5
6
6
6
6
6
6
6
6
×
+ ×
+ ×
+ ×
+ ×
+ ×
=
Also
E(X2) =
2
2
2
2
2
2
1
1
1
1
1
1
91
1
2
3
4
5
6
6
6
6
6
6
6
6
×
+
×
+
×
+
×
+
×
+
×
=
Thus,
Var (X) = E (X2) – (E(X))2
=
2
91
21
91
441
6
6
6
36
⎛
⎞
−
=
−
⎜
⎟
⎝
⎠
35
12
Example 29 Two cards are drawn simultaneously (or successively without replacement)
from a well shuffled pack of 52 cards Find the mean, variance and standard deviation
of the number of kings |
1 | 7404-7407 | Then X is a random variable
which can take values 1, 2, 3, 4, 5, or 6 © NCERT
not to be republished
568
MATHEMATICS
Also
P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1
6
Therefore, the Probability distribution of X is
X
1
2
3
4
5
6
P(X)
1
6
1
6
1
6
1
6
1
6
1
6
Now
E(X) =
1
(
)
n
i
i
i
x p x
=
1
1
1
1
1
1
21
1
2
3
4
5
6
6
6
6
6
6
6
6
×
+ ×
+ ×
+ ×
+ ×
+ ×
=
Also
E(X2) =
2
2
2
2
2
2
1
1
1
1
1
1
91
1
2
3
4
5
6
6
6
6
6
6
6
6
×
+
×
+
×
+
×
+
×
+
×
=
Thus,
Var (X) = E (X2) – (E(X))2
=
2
91
21
91
441
6
6
6
36
⎛
⎞
−
=
−
⎜
⎟
⎝
⎠
35
12
Example 29 Two cards are drawn simultaneously (or successively without replacement)
from a well shuffled pack of 52 cards Find the mean, variance and standard deviation
of the number of kings Solution Let X denote the number of kings in a draw of two cards |
1 | 7405-7408 | © NCERT
not to be republished
568
MATHEMATICS
Also
P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1
6
Therefore, the Probability distribution of X is
X
1
2
3
4
5
6
P(X)
1
6
1
6
1
6
1
6
1
6
1
6
Now
E(X) =
1
(
)
n
i
i
i
x p x
=
1
1
1
1
1
1
21
1
2
3
4
5
6
6
6
6
6
6
6
6
×
+ ×
+ ×
+ ×
+ ×
+ ×
=
Also
E(X2) =
2
2
2
2
2
2
1
1
1
1
1
1
91
1
2
3
4
5
6
6
6
6
6
6
6
6
×
+
×
+
×
+
×
+
×
+
×
=
Thus,
Var (X) = E (X2) – (E(X))2
=
2
91
21
91
441
6
6
6
36
⎛
⎞
−
=
−
⎜
⎟
⎝
⎠
35
12
Example 29 Two cards are drawn simultaneously (or successively without replacement)
from a well shuffled pack of 52 cards Find the mean, variance and standard deviation
of the number of kings Solution Let X denote the number of kings in a draw of two cards X is a random
variable which can assume the values 0, 1 or 2 |
1 | 7406-7409 | Find the mean, variance and standard deviation
of the number of kings Solution Let X denote the number of kings in a draw of two cards X is a random
variable which can assume the values 0, 1 or 2 Now
P(X = 0) = P (no king)
48
2
52
2
48 |
1 | 7407-7410 | Solution Let X denote the number of kings in a draw of two cards X is a random
variable which can assume the values 0, 1 or 2 Now
P(X = 0) = P (no king)
48
2
52
2
48 C
48 47
188
2 |
1 | 7408-7411 | X is a random
variable which can assume the values 0, 1 or 2 Now
P(X = 0) = P (no king)
48
2
52
2
48 C
48 47
188
2 (48
52 |
1 | 7409-7412 | Now
P(X = 0) = P (no king)
48
2
52
2
48 C
48 47
188
2 (48
52 2) |
1 | 7410-7413 | C
48 47
188
2 (48
52 2) 52 51
221
C
2 |
1 | 7411-7414 | (48
52 2) 52 51
221
C
2 (52
2) |
1 | 7412-7415 | 2) 52 51
221
C
2 (52
2) P(X = 1) = P (one king and one non-king)
4
148
1
52
2
C
CC
=
= 4 48 2
32
52 51
221
×
× =
×
© NCERT
not to be republished
PROBABILITY 569
and
P(X = 2) = P (two kings) =
4
2
52
2
C
4 3
1
52 51
221
C
×
=
=
×
Thus, the probability distribution of X is
X
0
1
2
P(X)
188
221
32
221
1
221
Now Mean of
X = E(X) =
1
(
)
n
i
i
i
x p x
=
188
32
1
34
0
1
2
221
221
221
221
×
+ ×
+ ×
=
Also
E(X2) =
2
1
(
)
n
i
i
i
x p x
=∑
=
2
2
2
188
32
1
36
0
1
2
221
221
221
221
×
+
×
+
×
=
Now
Var(X) = E(X2) – [E(X)]2
=
2
2
36
34
6800
221–
221
(221)
⎛
⎞ =
⎜
⎟
⎝
⎠
Therefore
σx =
6800
Var(X)
0 |
1 | 7413-7416 | 52 51
221
C
2 (52
2) P(X = 1) = P (one king and one non-king)
4
148
1
52
2
C
CC
=
= 4 48 2
32
52 51
221
×
× =
×
© NCERT
not to be republished
PROBABILITY 569
and
P(X = 2) = P (two kings) =
4
2
52
2
C
4 3
1
52 51
221
C
×
=
=
×
Thus, the probability distribution of X is
X
0
1
2
P(X)
188
221
32
221
1
221
Now Mean of
X = E(X) =
1
(
)
n
i
i
i
x p x
=
188
32
1
34
0
1
2
221
221
221
221
×
+ ×
+ ×
=
Also
E(X2) =
2
1
(
)
n
i
i
i
x p x
=∑
=
2
2
2
188
32
1
36
0
1
2
221
221
221
221
×
+
×
+
×
=
Now
Var(X) = E(X2) – [E(X)]2
=
2
2
36
34
6800
221–
221
(221)
⎛
⎞ =
⎜
⎟
⎝
⎠
Therefore
σx =
6800
Var(X)
0 37
221
=
=
EXERCISE 13 |
1 | 7414-7417 | (52
2) P(X = 1) = P (one king and one non-king)
4
148
1
52
2
C
CC
=
= 4 48 2
32
52 51
221
×
× =
×
© NCERT
not to be republished
PROBABILITY 569
and
P(X = 2) = P (two kings) =
4
2
52
2
C
4 3
1
52 51
221
C
×
=
=
×
Thus, the probability distribution of X is
X
0
1
2
P(X)
188
221
32
221
1
221
Now Mean of
X = E(X) =
1
(
)
n
i
i
i
x p x
=
188
32
1
34
0
1
2
221
221
221
221
×
+ ×
+ ×
=
Also
E(X2) =
2
1
(
)
n
i
i
i
x p x
=∑
=
2
2
2
188
32
1
36
0
1
2
221
221
221
221
×
+
×
+
×
=
Now
Var(X) = E(X2) – [E(X)]2
=
2
2
36
34
6800
221–
221
(221)
⎛
⎞ =
⎜
⎟
⎝
⎠
Therefore
σx =
6800
Var(X)
0 37
221
=
=
EXERCISE 13 4
1 |
1 | 7415-7418 | P(X = 1) = P (one king and one non-king)
4
148
1
52
2
C
CC
=
= 4 48 2
32
52 51
221
×
× =
×
© NCERT
not to be republished
PROBABILITY 569
and
P(X = 2) = P (two kings) =
4
2
52
2
C
4 3
1
52 51
221
C
×
=
=
×
Thus, the probability distribution of X is
X
0
1
2
P(X)
188
221
32
221
1
221
Now Mean of
X = E(X) =
1
(
)
n
i
i
i
x p x
=
188
32
1
34
0
1
2
221
221
221
221
×
+ ×
+ ×
=
Also
E(X2) =
2
1
(
)
n
i
i
i
x p x
=∑
=
2
2
2
188
32
1
36
0
1
2
221
221
221
221
×
+
×
+
×
=
Now
Var(X) = E(X2) – [E(X)]2
=
2
2
36
34
6800
221–
221
(221)
⎛
⎞ =
⎜
⎟
⎝
⎠
Therefore
σx =
6800
Var(X)
0 37
221
=
=
EXERCISE 13 4
1 State which of the following are not the probability distributions of a random
variable |
1 | 7416-7419 | 37
221
=
=
EXERCISE 13 4
1 State which of the following are not the probability distributions of a random
variable Give reasons for your answer |
1 | 7417-7420 | 4
1 State which of the following are not the probability distributions of a random
variable Give reasons for your answer (i)
X
0
1
2
P(X)
0 |
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