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1 | 7518-7521 | Clearly, six different cases are there as listed below:
SFFFFF, FSFFFF, FFSFFF, FFFSFF, FFFFSF, FFFFFS Similarly, two successes and four failures can have
6 4 2 |
1 | 7519-7522 | Similarly, two successes and four failures can have
6 4 2 combinations |
1 | 7520-7523 | 4 2 combinations It will be
lengthy job to list all of these ways |
1 | 7521-7524 | 2 combinations It will be
lengthy job to list all of these ways Therefore, calculation of probabilities of 0, 1, 2, |
1 | 7522-7525 | combinations It will be
lengthy job to list all of these ways Therefore, calculation of probabilities of 0, 1, 2, ,
n number of successes may be lengthy and time consuming |
1 | 7523-7526 | It will be
lengthy job to list all of these ways Therefore, calculation of probabilities of 0, 1, 2, ,
n number of successes may be lengthy and time consuming To avoid the lengthy
calculations and listing of all the possible cases, for the probabilities of number of
successes in n-Bernoulli trials, a formula is derived |
1 | 7524-7527 | Therefore, calculation of probabilities of 0, 1, 2, ,
n number of successes may be lengthy and time consuming To avoid the lengthy
calculations and listing of all the possible cases, for the probabilities of number of
successes in n-Bernoulli trials, a formula is derived For this purpose, let us take the
experiment made up of three Bernoulli trials with probabilities p and q = 1 – p for
success and failure respectively in each trial |
1 | 7525-7528 | ,
n number of successes may be lengthy and time consuming To avoid the lengthy
calculations and listing of all the possible cases, for the probabilities of number of
successes in n-Bernoulli trials, a formula is derived For this purpose, let us take the
experiment made up of three Bernoulli trials with probabilities p and q = 1 – p for
success and failure respectively in each trial The sample space of the experiment is
the set
S = {SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF}
The number of successes is a random variable X and can take values 0, 1, 2, or 3 |
1 | 7526-7529 | To avoid the lengthy
calculations and listing of all the possible cases, for the probabilities of number of
successes in n-Bernoulli trials, a formula is derived For this purpose, let us take the
experiment made up of three Bernoulli trials with probabilities p and q = 1 – p for
success and failure respectively in each trial The sample space of the experiment is
the set
S = {SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF}
The number of successes is a random variable X and can take values 0, 1, 2, or 3 The probability distribution of the number of successes is as below :
P(X = 0) = P(no success)
= P({FFF}) = P(F) P(F) P(F)
= q |
1 | 7527-7530 | For this purpose, let us take the
experiment made up of three Bernoulli trials with probabilities p and q = 1 – p for
success and failure respectively in each trial The sample space of the experiment is
the set
S = {SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF}
The number of successes is a random variable X and can take values 0, 1, 2, or 3 The probability distribution of the number of successes is as below :
P(X = 0) = P(no success)
= P({FFF}) = P(F) P(F) P(F)
= q q |
1 | 7528-7531 | The sample space of the experiment is
the set
S = {SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF}
The number of successes is a random variable X and can take values 0, 1, 2, or 3 The probability distribution of the number of successes is as below :
P(X = 0) = P(no success)
= P({FFF}) = P(F) P(F) P(F)
= q q q = q3 since the trials are independent
P(X = 1) = P(one successes)
= P({SFF, FSF, FFS})
= P({SFF}) + P({FSF}) + P({FFS})
= P(S) P(F) P(F) + P(F) P(S) P(F) + P(F) P(F) P(S)
= p |
1 | 7529-7532 | The probability distribution of the number of successes is as below :
P(X = 0) = P(no success)
= P({FFF}) = P(F) P(F) P(F)
= q q q = q3 since the trials are independent
P(X = 1) = P(one successes)
= P({SFF, FSF, FFS})
= P({SFF}) + P({FSF}) + P({FFS})
= P(S) P(F) P(F) + P(F) P(S) P(F) + P(F) P(F) P(S)
= p q |
1 | 7530-7533 | q q = q3 since the trials are independent
P(X = 1) = P(one successes)
= P({SFF, FSF, FFS})
= P({SFF}) + P({FSF}) + P({FFS})
= P(S) P(F) P(F) + P(F) P(S) P(F) + P(F) P(F) P(S)
= p q q + q |
1 | 7531-7534 | q = q3 since the trials are independent
P(X = 1) = P(one successes)
= P({SFF, FSF, FFS})
= P({SFF}) + P({FSF}) + P({FFS})
= P(S) P(F) P(F) + P(F) P(S) P(F) + P(F) P(F) P(S)
= p q q + q p |
1 | 7532-7535 | q q + q p q + q |
1 | 7533-7536 | q + q p q + q q |
1 | 7534-7537 | p q + q q p = 3pq2
P(X = 2) = P (two successes)
= P({SSF, SFS, FSS})
= P({SSF}) + P ({SFS}) + P({FSS})
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574
MATHEMATICS
= P(S) P(S) P(F) + P(S) P(F) P(S) + P(F) P(S) P(S)
= p |
1 | 7535-7538 | q + q q p = 3pq2
P(X = 2) = P (two successes)
= P({SSF, SFS, FSS})
= P({SSF}) + P ({SFS}) + P({FSS})
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574
MATHEMATICS
= P(S) P(S) P(F) + P(S) P(F) P(S) + P(F) P(S) P(S)
= p p |
1 | 7536-7539 | q p = 3pq2
P(X = 2) = P (two successes)
= P({SSF, SFS, FSS})
= P({SSF}) + P ({SFS}) + P({FSS})
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574
MATHEMATICS
= P(S) P(S) P(F) + P(S) P(F) P(S) + P(F) P(S) P(S)
= p p q |
1 | 7537-7540 | p = 3pq2
P(X = 2) = P (two successes)
= P({SSF, SFS, FSS})
= P({SSF}) + P ({SFS}) + P({FSS})
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574
MATHEMATICS
= P(S) P(S) P(F) + P(S) P(F) P(S) + P(F) P(S) P(S)
= p p q + p |
1 | 7538-7541 | p q + p q |
1 | 7539-7542 | q + p q p + q |
1 | 7540-7543 | + p q p + q p |
1 | 7541-7544 | q p + q p p = 3p2q
and
P(X = 3) = P(three success) = P ({SSS})
= P(S) |
1 | 7542-7545 | p + q p p = 3p2q
and
P(X = 3) = P(three success) = P ({SSS})
= P(S) P(S) |
1 | 7543-7546 | p p = 3p2q
and
P(X = 3) = P(three success) = P ({SSS})
= P(S) P(S) P(S) = p3
Thus, the probability distribution of X is
X
0
1
2
3
P(X)
q 3
3q2p
3qp2
p 3
Also, the binominal expansion of (q + p)3 is
q
q
p
qp
p
3
3 2
3
2
3
+
+
+
Note that the probabilities of 0, 1, 2 or 3 successes are respectively the 1st, 2nd,
3rd and 4th term in the expansion of (q + p)3 |
1 | 7544-7547 | p = 3p2q
and
P(X = 3) = P(three success) = P ({SSS})
= P(S) P(S) P(S) = p3
Thus, the probability distribution of X is
X
0
1
2
3
P(X)
q 3
3q2p
3qp2
p 3
Also, the binominal expansion of (q + p)3 is
q
q
p
qp
p
3
3 2
3
2
3
+
+
+
Note that the probabilities of 0, 1, 2 or 3 successes are respectively the 1st, 2nd,
3rd and 4th term in the expansion of (q + p)3 Also, since q + p = 1, it follows that the sum of these probabilities, as expected, is 1 |
1 | 7545-7548 | P(S) P(S) = p3
Thus, the probability distribution of X is
X
0
1
2
3
P(X)
q 3
3q2p
3qp2
p 3
Also, the binominal expansion of (q + p)3 is
q
q
p
qp
p
3
3 2
3
2
3
+
+
+
Note that the probabilities of 0, 1, 2 or 3 successes are respectively the 1st, 2nd,
3rd and 4th term in the expansion of (q + p)3 Also, since q + p = 1, it follows that the sum of these probabilities, as expected, is 1 Thus, we may conclude that in an experiment of n-Bernoulli trials, the probabilities
of 0, 1, 2, |
1 | 7546-7549 | P(S) = p3
Thus, the probability distribution of X is
X
0
1
2
3
P(X)
q 3
3q2p
3qp2
p 3
Also, the binominal expansion of (q + p)3 is
q
q
p
qp
p
3
3 2
3
2
3
+
+
+
Note that the probabilities of 0, 1, 2 or 3 successes are respectively the 1st, 2nd,
3rd and 4th term in the expansion of (q + p)3 Also, since q + p = 1, it follows that the sum of these probabilities, as expected, is 1 Thus, we may conclude that in an experiment of n-Bernoulli trials, the probabilities
of 0, 1, 2, , n successes can be obtained as 1st, 2nd, |
1 | 7547-7550 | Also, since q + p = 1, it follows that the sum of these probabilities, as expected, is 1 Thus, we may conclude that in an experiment of n-Bernoulli trials, the probabilities
of 0, 1, 2, , n successes can be obtained as 1st, 2nd, ,(n + 1)th terms in the expansion
of (q + p)n |
1 | 7548-7551 | Thus, we may conclude that in an experiment of n-Bernoulli trials, the probabilities
of 0, 1, 2, , n successes can be obtained as 1st, 2nd, ,(n + 1)th terms in the expansion
of (q + p)n To prove this assertion (result), let us find the probability of x-successes in
an experiment of n-Bernoulli trials |
1 | 7549-7552 | , n successes can be obtained as 1st, 2nd, ,(n + 1)th terms in the expansion
of (q + p)n To prove this assertion (result), let us find the probability of x-successes in
an experiment of n-Bernoulli trials Clearly, in case of x successes (S), there will be (n – x) failures (F) |
1 | 7550-7553 | ,(n + 1)th terms in the expansion
of (q + p)n To prove this assertion (result), let us find the probability of x-successes in
an experiment of n-Bernoulli trials Clearly, in case of x successes (S), there will be (n – x) failures (F) Now, x successes (S) and (n – x) failures (F) can be obtained in |
1 | 7551-7554 | To prove this assertion (result), let us find the probability of x-successes in
an experiment of n-Bernoulli trials Clearly, in case of x successes (S), there will be (n – x) failures (F) Now, x successes (S) and (n – x) failures (F) can be obtained in ( |
1 | 7552-7555 | Clearly, in case of x successes (S), there will be (n – x) failures (F) Now, x successes (S) and (n – x) failures (F) can be obtained in ( ) |
1 | 7553-7556 | Now, x successes (S) and (n – x) failures (F) can be obtained in ( ) n
x n
x
ways |
1 | 7554-7557 | ( ) n
x n
x
ways In each of these ways, the probability of x successes and (n − x) failures is
= P(x successes) |
1 | 7555-7558 | ) n
x n
x
ways In each of these ways, the probability of x successes and (n − x) failures is
= P(x successes) P(n–x) failures is
=
times
(
) times
P(S) |
1 | 7556-7559 | n
x n
x
ways In each of these ways, the probability of x successes and (n − x) failures is
= P(x successes) P(n–x) failures is
=
times
(
) times
P(S) P(S) |
1 | 7557-7560 | In each of these ways, the probability of x successes and (n − x) failures is
= P(x successes) P(n–x) failures is
=
times
(
) times
P(S) P(S) P(S)
P(F) |
1 | 7558-7561 | P(n–x) failures is
=
times
(
) times
P(S) P(S) P(S)
P(F) P(F) |
1 | 7559-7562 | P(S) P(S)
P(F) P(F) P(F)
x
n x
1442443
1442443 = px qn–x
Thus, the probability of x successes in n-Bernoulli trials is |
1 | 7560-7563 | P(S)
P(F) P(F) P(F)
x
n x
1442443
1442443 = px qn–x
Thus, the probability of x successes in n-Bernoulli trials is ( |
1 | 7561-7564 | P(F) P(F)
x
n x
1442443
1442443 = px qn–x
Thus, the probability of x successes in n-Bernoulli trials is ( ) |
1 | 7562-7565 | P(F)
x
n x
1442443
1442443 = px qn–x
Thus, the probability of x successes in n-Bernoulli trials is ( ) n
x n
−x
px qn–x
or nCx
px qn–x
Thus
P(x successes) =
nC
x
n x
x p q − , x = 0, 1, 2, |
1 | 7563-7566 | ( ) n
x n
−x
px qn–x
or nCx
px qn–x
Thus
P(x successes) =
nC
x
n x
x p q − , x = 0, 1, 2, ,n |
1 | 7564-7567 | ) n
x n
−x
px qn–x
or nCx
px qn–x
Thus
P(x successes) =
nC
x
n x
x p q − , x = 0, 1, 2, ,n (q = 1 – p)
Clearly, P(x successes), i |
1 | 7565-7568 | n
x n
−x
px qn–x
or nCx
px qn–x
Thus
P(x successes) =
nC
x
n x
x p q − , x = 0, 1, 2, ,n (q = 1 – p)
Clearly, P(x successes), i e |
1 | 7566-7569 | ,n (q = 1 – p)
Clearly, P(x successes), i e C
n
x
n x
x p q − is the (x + 1)th term in the binomial
expansion of (q + p)n |
1 | 7567-7570 | (q = 1 – p)
Clearly, P(x successes), i e C
n
x
n x
x p q − is the (x + 1)th term in the binomial
expansion of (q + p)n Thus, the probability distribution of number of successes in an experiment consisting
of n Bernoulli trials may be obtained by the binomial expansion of (q + p)n |
1 | 7568-7571 | e C
n
x
n x
x p q − is the (x + 1)th term in the binomial
expansion of (q + p)n Thus, the probability distribution of number of successes in an experiment consisting
of n Bernoulli trials may be obtained by the binomial expansion of (q + p)n Hence, this
© NCERT
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PROBABILITY 575
distribution of number of successes X can be written as
X
0
1
2 |
1 | 7569-7572 | C
n
x
n x
x p q − is the (x + 1)th term in the binomial
expansion of (q + p)n Thus, the probability distribution of number of successes in an experiment consisting
of n Bernoulli trials may be obtained by the binomial expansion of (q + p)n Hence, this
© NCERT
not to be republished
PROBABILITY 575
distribution of number of successes X can be written as
X
0
1
2 x |
1 | 7570-7573 | Thus, the probability distribution of number of successes in an experiment consisting
of n Bernoulli trials may be obtained by the binomial expansion of (q + p)n Hence, this
© NCERT
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PROBABILITY 575
distribution of number of successes X can be written as
X
0
1
2 x n
P(X)
nC0 qn
nC1 qn–1p1
nC2 qn–2p2
nCx qn–xpx
nCn pn
The above probability distribution is known as binomial distribution with parameters
n and p, because for given values of n and p, we can find the complete probability
distribution |
1 | 7571-7574 | Hence, this
© NCERT
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PROBABILITY 575
distribution of number of successes X can be written as
X
0
1
2 x n
P(X)
nC0 qn
nC1 qn–1p1
nC2 qn–2p2
nCx qn–xpx
nCn pn
The above probability distribution is known as binomial distribution with parameters
n and p, because for given values of n and p, we can find the complete probability
distribution The probability of x successes P(X = x) is also denoted by P(x) and is given by
P(x) = nCx qn–xpx, x = 0, 1, |
1 | 7572-7575 | x n
P(X)
nC0 qn
nC1 qn–1p1
nC2 qn–2p2
nCx qn–xpx
nCn pn
The above probability distribution is known as binomial distribution with parameters
n and p, because for given values of n and p, we can find the complete probability
distribution The probability of x successes P(X = x) is also denoted by P(x) and is given by
P(x) = nCx qn–xpx, x = 0, 1, , n |
1 | 7573-7576 | n
P(X)
nC0 qn
nC1 qn–1p1
nC2 qn–2p2
nCx qn–xpx
nCn pn
The above probability distribution is known as binomial distribution with parameters
n and p, because for given values of n and p, we can find the complete probability
distribution The probability of x successes P(X = x) is also denoted by P(x) and is given by
P(x) = nCx qn–xpx, x = 0, 1, , n (q = 1 – p)
This P(x) is called the probability function of the binomial distribution |
1 | 7574-7577 | The probability of x successes P(X = x) is also denoted by P(x) and is given by
P(x) = nCx qn–xpx, x = 0, 1, , n (q = 1 – p)
This P(x) is called the probability function of the binomial distribution A binomial distribution with n-Bernoulli trials and probability of success in each
trial as p, is denoted by B(n, p) |
1 | 7575-7578 | , n (q = 1 – p)
This P(x) is called the probability function of the binomial distribution A binomial distribution with n-Bernoulli trials and probability of success in each
trial as p, is denoted by B(n, p) Let us now take up some examples |
1 | 7576-7579 | (q = 1 – p)
This P(x) is called the probability function of the binomial distribution A binomial distribution with n-Bernoulli trials and probability of success in each
trial as p, is denoted by B(n, p) Let us now take up some examples Example 31 If a fair coin is tossed 10 times, find the probability of
(i)
exactly six heads
(ii)
at least six heads
(iii)
at most six heads
Solution The repeated tosses of a coin are Bernoulli trials |
1 | 7577-7580 | A binomial distribution with n-Bernoulli trials and probability of success in each
trial as p, is denoted by B(n, p) Let us now take up some examples Example 31 If a fair coin is tossed 10 times, find the probability of
(i)
exactly six heads
(ii)
at least six heads
(iii)
at most six heads
Solution The repeated tosses of a coin are Bernoulli trials Let X denote the number
of heads in an experiment of 10 trials |
1 | 7578-7581 | Let us now take up some examples Example 31 If a fair coin is tossed 10 times, find the probability of
(i)
exactly six heads
(ii)
at least six heads
(iii)
at most six heads
Solution The repeated tosses of a coin are Bernoulli trials Let X denote the number
of heads in an experiment of 10 trials Clearly, X has the binomial distribution with n = 10 and p = 1
2
Therefore
P(X = x) = nCxqn–xpx, x = 0, 1, 2, |
1 | 7579-7582 | Example 31 If a fair coin is tossed 10 times, find the probability of
(i)
exactly six heads
(ii)
at least six heads
(iii)
at most six heads
Solution The repeated tosses of a coin are Bernoulli trials Let X denote the number
of heads in an experiment of 10 trials Clearly, X has the binomial distribution with n = 10 and p = 1
2
Therefore
P(X = x) = nCxqn–xpx, x = 0, 1, 2, ,n
Here
n = 10,
21
p
, q = 1 – p = 1
2
Therefore
P(X = x) =
10
10
10
10
1
1
1
C
C
2
2
2
x
x
x
x
−
⎛
⎞
⎛
⎞
⎛
⎞
=
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
Now
(i) P(X = 6) =
10
10
6
10
1
10 |
1 | 7580-7583 | Let X denote the number
of heads in an experiment of 10 trials Clearly, X has the binomial distribution with n = 10 and p = 1
2
Therefore
P(X = x) = nCxqn–xpx, x = 0, 1, 2, ,n
Here
n = 10,
21
p
, q = 1 – p = 1
2
Therefore
P(X = x) =
10
10
10
10
1
1
1
C
C
2
2
2
x
x
x
x
−
⎛
⎞
⎛
⎞
⎛
⎞
=
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
Now
(i) P(X = 6) =
10
10
6
10
1
10 1
105
C
2
6 |
1 | 7581-7584 | Clearly, X has the binomial distribution with n = 10 and p = 1
2
Therefore
P(X = x) = nCxqn–xpx, x = 0, 1, 2, ,n
Here
n = 10,
21
p
, q = 1 – p = 1
2
Therefore
P(X = x) =
10
10
10
10
1
1
1
C
C
2
2
2
x
x
x
x
−
⎛
⎞
⎛
⎞
⎛
⎞
=
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
Now
(i) P(X = 6) =
10
10
6
10
1
10 1
105
C
2
6 4 |
1 | 7582-7585 | ,n
Here
n = 10,
21
p
, q = 1 – p = 1
2
Therefore
P(X = x) =
10
10
10
10
1
1
1
C
C
2
2
2
x
x
x
x
−
⎛
⎞
⎛
⎞
⎛
⎞
=
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
Now
(i) P(X = 6) =
10
10
6
10
1
10 1
105
C
2
6 4 512
2
⎛
⎞
=
=
⎜
⎟
×
⎝
⎠
(ii) P(at least six heads) = P(X ≥ 6)
= P (X = 6) + P (X = 7) + P (X = 8) + P(X = 9) + P (X = 10)
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576
MATHEMATICS
=
10
10
10
10
10
10
10
10
10
10
6
7
8
9
10
1
1
1
1
1
C
C
C
C
C
2
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
=
10
10 |
1 | 7583-7586 | 1
105
C
2
6 4 512
2
⎛
⎞
=
=
⎜
⎟
×
⎝
⎠
(ii) P(at least six heads) = P(X ≥ 6)
= P (X = 6) + P (X = 7) + P (X = 8) + P(X = 9) + P (X = 10)
© NCERT
not to be republished
576
MATHEMATICS
=
10
10
10
10
10
10
10
10
10
10
6
7
8
9
10
1
1
1
1
1
C
C
C
C
C
2
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
=
10
10 10 |
1 | 7584-7587 | 4 512
2
⎛
⎞
=
=
⎜
⎟
×
⎝
⎠
(ii) P(at least six heads) = P(X ≥ 6)
= P (X = 6) + P (X = 7) + P (X = 8) + P(X = 9) + P (X = 10)
© NCERT
not to be republished
576
MATHEMATICS
=
10
10
10
10
10
10
10
10
10
10
6
7
8
9
10
1
1
1
1
1
C
C
C
C
C
2
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
=
10
10 10 10 |
1 | 7585-7588 | 512
2
⎛
⎞
=
=
⎜
⎟
×
⎝
⎠
(ii) P(at least six heads) = P(X ≥ 6)
= P (X = 6) + P (X = 7) + P (X = 8) + P(X = 9) + P (X = 10)
© NCERT
not to be republished
576
MATHEMATICS
=
10
10
10
10
10
10
10
10
10
10
6
7
8
9
10
1
1
1
1
1
C
C
C
C
C
2
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
=
10
10 10 10 10 |
1 | 7586-7589 | 10 10 10 10 |
1 | 7587-7590 | 10 10 10 1
6 |
1 | 7588-7591 | 10 10 1
6 4 |
1 | 7589-7592 | 10 1
6 4 7 |
1 | 7590-7593 | 1
6 4 7 3 |
1 | 7591-7594 | 4 7 3 8 |
1 | 7592-7595 | 7 3 8 2 |
1 | 7593-7596 | 3 8 2 9 |
1 | 7594-7597 | 8 2 9 1 |
1 | 7595-7598 | 2 9 1 10 |
1 | 7596-7599 | 9 1 10 2
193
512
=
(iii) P(at most six heads) = P(X ≤ 6)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
+ P (X = 4) + P (X = 5) + P (X = 6)
=
10
10
10
10
10
10
10
1
2
3
1
1
1
1
C
C
C
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
+
10
10
10
10
10
10
4
5
6
1
1
1
C
C
C
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
= 848
53
1024
64
=
Example 32 Ten eggs are drawn successively with replacement from a lot containing
10% defective eggs |
1 | 7597-7600 | 1 10 2
193
512
=
(iii) P(at most six heads) = P(X ≤ 6)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
+ P (X = 4) + P (X = 5) + P (X = 6)
=
10
10
10
10
10
10
10
1
2
3
1
1
1
1
C
C
C
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
+
10
10
10
10
10
10
4
5
6
1
1
1
C
C
C
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
= 848
53
1024
64
=
Example 32 Ten eggs are drawn successively with replacement from a lot containing
10% defective eggs Find the probability that there is at least one defective egg |
1 | 7598-7601 | 10 2
193
512
=
(iii) P(at most six heads) = P(X ≤ 6)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
+ P (X = 4) + P (X = 5) + P (X = 6)
=
10
10
10
10
10
10
10
1
2
3
1
1
1
1
C
C
C
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
+
10
10
10
10
10
10
4
5
6
1
1
1
C
C
C
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
= 848
53
1024
64
=
Example 32 Ten eggs are drawn successively with replacement from a lot containing
10% defective eggs Find the probability that there is at least one defective egg Solution Let X denote the number of defective eggs in the 10 eggs drawn |
1 | 7599-7602 | 2
193
512
=
(iii) P(at most six heads) = P(X ≤ 6)
= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
+ P (X = 4) + P (X = 5) + P (X = 6)
=
10
10
10
10
10
10
10
1
2
3
1
1
1
1
C
C
C
2
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
+
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
+
10
10
10
10
10
10
4
5
6
1
1
1
C
C
C
2
2
2
⎛
⎞
⎛
⎞
⎛
⎞
+
+
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
= 848
53
1024
64
=
Example 32 Ten eggs are drawn successively with replacement from a lot containing
10% defective eggs Find the probability that there is at least one defective egg Solution Let X denote the number of defective eggs in the 10 eggs drawn Since the
drawing is done with replacement, the trials are Bernoulli trials |
1 | 7600-7603 | Find the probability that there is at least one defective egg Solution Let X denote the number of defective eggs in the 10 eggs drawn Since the
drawing is done with replacement, the trials are Bernoulli trials Clearly, X has the
binomial distribution with n = 10 and
10
1
100
10
p |
1 | 7601-7604 | Solution Let X denote the number of defective eggs in the 10 eggs drawn Since the
drawing is done with replacement, the trials are Bernoulli trials Clearly, X has the
binomial distribution with n = 10 and
10
1
100
10
p Therefore
q =
9
1
−p10
=
Now
P(at least one defective egg) = P(X ≥ 1) = 1 – P (X = 0)
=
10
10
0
9
1
C
⎛10
⎞
−
⎜
⎟
⎝
⎠ =
10
910
1
10
−
EXERCISE 13 |
1 | 7602-7605 | Since the
drawing is done with replacement, the trials are Bernoulli trials Clearly, X has the
binomial distribution with n = 10 and
10
1
100
10
p Therefore
q =
9
1
−p10
=
Now
P(at least one defective egg) = P(X ≥ 1) = 1 – P (X = 0)
=
10
10
0
9
1
C
⎛10
⎞
−
⎜
⎟
⎝
⎠ =
10
910
1
10
−
EXERCISE 13 5
1 |
1 | 7603-7606 | Clearly, X has the
binomial distribution with n = 10 and
10
1
100
10
p Therefore
q =
9
1
−p10
=
Now
P(at least one defective egg) = P(X ≥ 1) = 1 – P (X = 0)
=
10
10
0
9
1
C
⎛10
⎞
−
⎜
⎟
⎝
⎠ =
10
910
1
10
−
EXERCISE 13 5
1 A die is thrown 6 times |
1 | 7604-7607 | Therefore
q =
9
1
−p10
=
Now
P(at least one defective egg) = P(X ≥ 1) = 1 – P (X = 0)
=
10
10
0
9
1
C
⎛10
⎞
−
⎜
⎟
⎝
⎠ =
10
910
1
10
−
EXERCISE 13 5
1 A die is thrown 6 times If ‘getting an odd number’ is a success, what is the
probability of
(i) 5 successes |
1 | 7605-7608 | 5
1 A die is thrown 6 times If ‘getting an odd number’ is a success, what is the
probability of
(i) 5 successes (ii) at least 5 successes |
1 | 7606-7609 | A die is thrown 6 times If ‘getting an odd number’ is a success, what is the
probability of
(i) 5 successes (ii) at least 5 successes (iii) at most 5 successes |
1 | 7607-7610 | If ‘getting an odd number’ is a success, what is the
probability of
(i) 5 successes (ii) at least 5 successes (iii) at most 5 successes © NCERT
not to be republished
PROBABILITY 577
2 |
1 | 7608-7611 | (ii) at least 5 successes (iii) at most 5 successes © NCERT
not to be republished
PROBABILITY 577
2 A pair of dice is thrown 4 times |
1 | 7609-7612 | (iii) at most 5 successes © NCERT
not to be republished
PROBABILITY 577
2 A pair of dice is thrown 4 times If getting a doublet is considered a success, find
the probability of two successes |
1 | 7610-7613 | © NCERT
not to be republished
PROBABILITY 577
2 A pair of dice is thrown 4 times If getting a doublet is considered a success, find
the probability of two successes 3 |
1 | 7611-7614 | A pair of dice is thrown 4 times If getting a doublet is considered a success, find
the probability of two successes 3 There are 5% defective items in a large bulk of items |
1 | 7612-7615 | If getting a doublet is considered a success, find
the probability of two successes 3 There are 5% defective items in a large bulk of items What is the probability
that a sample of 10 items will include not more than one defective item |
1 | 7613-7616 | 3 There are 5% defective items in a large bulk of items What is the probability
that a sample of 10 items will include not more than one defective item 4 |
1 | 7614-7617 | There are 5% defective items in a large bulk of items What is the probability
that a sample of 10 items will include not more than one defective item 4 Five cards are drawn successively with replacement from a well-shuffled deck
of 52 cards |
1 | 7615-7618 | What is the probability
that a sample of 10 items will include not more than one defective item 4 Five cards are drawn successively with replacement from a well-shuffled deck
of 52 cards What is the probability that
(i) all the five cards are spades |
1 | 7616-7619 | 4 Five cards are drawn successively with replacement from a well-shuffled deck
of 52 cards What is the probability that
(i) all the five cards are spades (ii) only 3 cards are spades |
1 | 7617-7620 | Five cards are drawn successively with replacement from a well-shuffled deck
of 52 cards What is the probability that
(i) all the five cards are spades (ii) only 3 cards are spades (iii) none is a spade |
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