Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 7118-7121 | © NCERT
not to be republished
552
MATHEMATICS
Solution Let E1, E2 and E3 be the events that boxes I, II and III are chosen, respectively Then
P(E1) = P(E2) = P(E3) = 1
3
Also, let A be the event that ‘the coin drawn is of gold’
Then
P(A|E1) = P(a gold coin from bag I) = 2
2 = 1
P(A|E2) = P(a gold coin from bag II) = 0
P(A|E3) = P(a gold coin from bag III) = 1
2
Now, the probability that the other coin in the box is of gold
= the probability that gold coin is drawn from the box I = P(E1|A)
By Bayes' theorem, we know that
P(E1|A) =
1
1
1
1
2
2
3
3
P(E )P(A|E )
P(E )P(A|E )+P(E )P(A|E )+P(E )P(A|E )
=
1 1
2
3
1
1
1
1
3
1
0
3
3
3
2
×
=
× + × + ×
Example 18 Suppose that the reliability of a HIV test is specified as follows:
Of people having HIV, 90% of the test detect the disease but 10% go undetected Of
people free of HIV, 99% of the test are judged HIV–ive but 1% are diagnosed as
showing HIV+ive |
1 | 7119-7122 | Then
P(E1) = P(E2) = P(E3) = 1
3
Also, let A be the event that ‘the coin drawn is of gold’
Then
P(A|E1) = P(a gold coin from bag I) = 2
2 = 1
P(A|E2) = P(a gold coin from bag II) = 0
P(A|E3) = P(a gold coin from bag III) = 1
2
Now, the probability that the other coin in the box is of gold
= the probability that gold coin is drawn from the box I = P(E1|A)
By Bayes' theorem, we know that
P(E1|A) =
1
1
1
1
2
2
3
3
P(E )P(A|E )
P(E )P(A|E )+P(E )P(A|E )+P(E )P(A|E )
=
1 1
2
3
1
1
1
1
3
1
0
3
3
3
2
×
=
× + × + ×
Example 18 Suppose that the reliability of a HIV test is specified as follows:
Of people having HIV, 90% of the test detect the disease but 10% go undetected Of
people free of HIV, 99% of the test are judged HIV–ive but 1% are diagnosed as
showing HIV+ive From a large population of which only 0 |
1 | 7120-7123 | = P(E1|A)
By Bayes' theorem, we know that
P(E1|A) =
1
1
1
1
2
2
3
3
P(E )P(A|E )
P(E )P(A|E )+P(E )P(A|E )+P(E )P(A|E )
=
1 1
2
3
1
1
1
1
3
1
0
3
3
3
2
×
=
× + × + ×
Example 18 Suppose that the reliability of a HIV test is specified as follows:
Of people having HIV, 90% of the test detect the disease but 10% go undetected Of
people free of HIV, 99% of the test are judged HIV–ive but 1% are diagnosed as
showing HIV+ive From a large population of which only 0 1% have HIV, one person
is selected at random, given the HIV test, and the pathologist reports him/her as
HIV+ive |
1 | 7121-7124 | Of
people free of HIV, 99% of the test are judged HIV–ive but 1% are diagnosed as
showing HIV+ive From a large population of which only 0 1% have HIV, one person
is selected at random, given the HIV test, and the pathologist reports him/her as
HIV+ive What is the probability that the person actually has HIV |
1 | 7122-7125 | From a large population of which only 0 1% have HIV, one person
is selected at random, given the HIV test, and the pathologist reports him/her as
HIV+ive What is the probability that the person actually has HIV Solution Let E denote the event that the person selected is actually having HIV and A
the event that the person's HIV test is diagnosed as +ive |
1 | 7123-7126 | 1% have HIV, one person
is selected at random, given the HIV test, and the pathologist reports him/her as
HIV+ive What is the probability that the person actually has HIV Solution Let E denote the event that the person selected is actually having HIV and A
the event that the person's HIV test is diagnosed as +ive We need to find P(E|A) |
1 | 7124-7127 | What is the probability that the person actually has HIV Solution Let E denote the event that the person selected is actually having HIV and A
the event that the person's HIV test is diagnosed as +ive We need to find P(E|A) Also E′ denotes the event that the person selected is actually not having HIV |
1 | 7125-7128 | Solution Let E denote the event that the person selected is actually having HIV and A
the event that the person's HIV test is diagnosed as +ive We need to find P(E|A) Also E′ denotes the event that the person selected is actually not having HIV Clearly, {E, E′} is a partition of the sample space of all people in the population |
1 | 7126-7129 | We need to find P(E|A) Also E′ denotes the event that the person selected is actually not having HIV Clearly, {E, E′} is a partition of the sample space of all people in the population We are given that
P(E) = 0 |
1 | 7127-7130 | Also E′ denotes the event that the person selected is actually not having HIV Clearly, {E, E′} is a partition of the sample space of all people in the population We are given that
P(E) = 0 1%
0 |
1 | 7128-7131 | Clearly, {E, E′} is a partition of the sample space of all people in the population We are given that
P(E) = 0 1%
0 1
0 |
1 | 7129-7132 | We are given that
P(E) = 0 1%
0 1
0 001
100
© NCERT
not to be republished
PROBABILITY 553
P(E′) = 1 – P(E) = 0 |
1 | 7130-7133 | 1%
0 1
0 001
100
© NCERT
not to be republished
PROBABILITY 553
P(E′) = 1 – P(E) = 0 999
P(A|E) = P(Person tested as HIV+ive given that he/she
is actually having HIV)
= 90%
90
0 |
1 | 7131-7134 | 1
0 001
100
© NCERT
not to be republished
PROBABILITY 553
P(E′) = 1 – P(E) = 0 999
P(A|E) = P(Person tested as HIV+ive given that he/she
is actually having HIV)
= 90%
90
0 9
100
and
P(A|E′) = P(Person tested as HIV +ive given that he/she
is actually not having HIV)
= 1% = 1
100 = 0 |
1 | 7132-7135 | 001
100
© NCERT
not to be republished
PROBABILITY 553
P(E′) = 1 – P(E) = 0 999
P(A|E) = P(Person tested as HIV+ive given that he/she
is actually having HIV)
= 90%
90
0 9
100
and
P(A|E′) = P(Person tested as HIV +ive given that he/she
is actually not having HIV)
= 1% = 1
100 = 0 01
Now, by Bayes' theorem
P(E|A) =
P(E)P(A|E)
P(E)P(A|E)+P(E )P(A|E )
=
0 |
1 | 7133-7136 | 999
P(A|E) = P(Person tested as HIV+ive given that he/she
is actually having HIV)
= 90%
90
0 9
100
and
P(A|E′) = P(Person tested as HIV +ive given that he/she
is actually not having HIV)
= 1% = 1
100 = 0 01
Now, by Bayes' theorem
P(E|A) =
P(E)P(A|E)
P(E)P(A|E)+P(E )P(A|E )
=
0 001 0 |
1 | 7134-7137 | 9
100
and
P(A|E′) = P(Person tested as HIV +ive given that he/she
is actually not having HIV)
= 1% = 1
100 = 0 01
Now, by Bayes' theorem
P(E|A) =
P(E)P(A|E)
P(E)P(A|E)+P(E )P(A|E )
=
0 001 0 9
90
0 |
1 | 7135-7138 | 01
Now, by Bayes' theorem
P(E|A) =
P(E)P(A|E)
P(E)P(A|E)+P(E )P(A|E )
=
0 001 0 9
90
0 001 0 |
1 | 7136-7139 | 001 0 9
90
0 001 0 9 0 |
1 | 7137-7140 | 9
90
0 001 0 9 0 999 0 |
1 | 7138-7141 | 001 0 9 0 999 0 01
1089
×
=
×
+
×
= 0 |
1 | 7139-7142 | 9 0 999 0 01
1089
×
=
×
+
×
= 0 083 approx |
1 | 7140-7143 | 999 0 01
1089
×
=
×
+
×
= 0 083 approx Thus, the probability that a person selected at random is actually having HIV
given that he/she is tested HIV+ive is 0 |
1 | 7141-7144 | 01
1089
×
=
×
+
×
= 0 083 approx Thus, the probability that a person selected at random is actually having HIV
given that he/she is tested HIV+ive is 0 083 |
1 | 7142-7145 | 083 approx Thus, the probability that a person selected at random is actually having HIV
given that he/she is tested HIV+ive is 0 083 Example 19 In a factory which manufactures bolts, machines A, B and C manufacture
respectively 25%, 35% and 40% of the bolts |
1 | 7143-7146 | Thus, the probability that a person selected at random is actually having HIV
given that he/she is tested HIV+ive is 0 083 Example 19 In a factory which manufactures bolts, machines A, B and C manufacture
respectively 25%, 35% and 40% of the bolts Of their outputs, 5, 4 and 2 percent are
respectively defective bolts |
1 | 7144-7147 | 083 Example 19 In a factory which manufactures bolts, machines A, B and C manufacture
respectively 25%, 35% and 40% of the bolts Of their outputs, 5, 4 and 2 percent are
respectively defective bolts A bolt is drawn at random from the product and is found
to be defective |
1 | 7145-7148 | Example 19 In a factory which manufactures bolts, machines A, B and C manufacture
respectively 25%, 35% and 40% of the bolts Of their outputs, 5, 4 and 2 percent are
respectively defective bolts A bolt is drawn at random from the product and is found
to be defective What is the probability that it is manufactured by the machine B |
1 | 7146-7149 | Of their outputs, 5, 4 and 2 percent are
respectively defective bolts A bolt is drawn at random from the product and is found
to be defective What is the probability that it is manufactured by the machine B Solution Let events B1, B2, B3 be the following :
B1 : the bolt is manufactured by machine A
B2 : the bolt is manufactured by machine B
B3 : the bolt is manufactured by machine C
Clearly, B1, B2, B3 are mutually exclusive and exhaustive events and hence, they
represent a partition of the sample space |
1 | 7147-7150 | A bolt is drawn at random from the product and is found
to be defective What is the probability that it is manufactured by the machine B Solution Let events B1, B2, B3 be the following :
B1 : the bolt is manufactured by machine A
B2 : the bolt is manufactured by machine B
B3 : the bolt is manufactured by machine C
Clearly, B1, B2, B3 are mutually exclusive and exhaustive events and hence, they
represent a partition of the sample space Let the event E be ‘the bolt is defective’ |
1 | 7148-7151 | What is the probability that it is manufactured by the machine B Solution Let events B1, B2, B3 be the following :
B1 : the bolt is manufactured by machine A
B2 : the bolt is manufactured by machine B
B3 : the bolt is manufactured by machine C
Clearly, B1, B2, B3 are mutually exclusive and exhaustive events and hence, they
represent a partition of the sample space Let the event E be ‘the bolt is defective’ The event E occurs with B1 or with B2 or with B3 |
1 | 7149-7152 | Solution Let events B1, B2, B3 be the following :
B1 : the bolt is manufactured by machine A
B2 : the bolt is manufactured by machine B
B3 : the bolt is manufactured by machine C
Clearly, B1, B2, B3 are mutually exclusive and exhaustive events and hence, they
represent a partition of the sample space Let the event E be ‘the bolt is defective’ The event E occurs with B1 or with B2 or with B3 Given that,
P(B1) = 25% = 0 |
1 | 7150-7153 | Let the event E be ‘the bolt is defective’ The event E occurs with B1 or with B2 or with B3 Given that,
P(B1) = 25% = 0 25, P (B2) = 0 |
1 | 7151-7154 | The event E occurs with B1 or with B2 or with B3 Given that,
P(B1) = 25% = 0 25, P (B2) = 0 35 and P(B3) = 0 |
1 | 7152-7155 | Given that,
P(B1) = 25% = 0 25, P (B2) = 0 35 and P(B3) = 0 40
Again P(E|B1) = Probability that the bolt drawn is defective given that it is manu-
factured by machine A = 5% = 0 |
1 | 7153-7156 | 25, P (B2) = 0 35 and P(B3) = 0 40
Again P(E|B1) = Probability that the bolt drawn is defective given that it is manu-
factured by machine A = 5% = 0 05
Similarly,
P(E|B2) = 0 |
1 | 7154-7157 | 35 and P(B3) = 0 40
Again P(E|B1) = Probability that the bolt drawn is defective given that it is manu-
factured by machine A = 5% = 0 05
Similarly,
P(E|B2) = 0 04, P(E|B3) = 0 |
1 | 7155-7158 | 40
Again P(E|B1) = Probability that the bolt drawn is defective given that it is manu-
factured by machine A = 5% = 0 05
Similarly,
P(E|B2) = 0 04, P(E|B3) = 0 02 |
1 | 7156-7159 | 05
Similarly,
P(E|B2) = 0 04, P(E|B3) = 0 02 © NCERT
not to be republished
554
MATHEMATICS
Hence, by Bayes' Theorem, we have
P(B2|E) =
2
2
1
1
2
2
3
3
P(B )P(E|B )
P(B )P(E|B )+P(B )P(E|B )+P(B )P(E|B )
=
0 |
1 | 7157-7160 | 04, P(E|B3) = 0 02 © NCERT
not to be republished
554
MATHEMATICS
Hence, by Bayes' Theorem, we have
P(B2|E) =
2
2
1
1
2
2
3
3
P(B )P(E|B )
P(B )P(E|B )+P(B )P(E|B )+P(B )P(E|B )
=
0 35
0 |
1 | 7158-7161 | 02 © NCERT
not to be republished
554
MATHEMATICS
Hence, by Bayes' Theorem, we have
P(B2|E) =
2
2
1
1
2
2
3
3
P(B )P(E|B )
P(B )P(E|B )+P(B )P(E|B )+P(B )P(E|B )
=
0 35
0 04
0 |
1 | 7159-7162 | © NCERT
not to be republished
554
MATHEMATICS
Hence, by Bayes' Theorem, we have
P(B2|E) =
2
2
1
1
2
2
3
3
P(B )P(E|B )
P(B )P(E|B )+P(B )P(E|B )+P(B )P(E|B )
=
0 35
0 04
0 25
0 |
1 | 7160-7163 | 35
0 04
0 25
0 05
0 |
1 | 7161-7164 | 04
0 25
0 05
0 35
0 |
1 | 7162-7165 | 25
0 05
0 35
0 04
0 |
1 | 7163-7166 | 05
0 35
0 04
0 40
0 |
1 | 7164-7167 | 35
0 04
0 40
0 02
×
×
+
×
+
×
= 0 |
1 | 7165-7168 | 04
0 40
0 02
×
×
+
×
+
×
= 0 0140
28
0 |
1 | 7166-7169 | 40
0 02
×
×
+
×
+
×
= 0 0140
28
0 0345
69
=
Example 20 A doctor is to visit a patient |
1 | 7167-7170 | 02
×
×
+
×
+
×
= 0 0140
28
0 0345
69
=
Example 20 A doctor is to visit a patient From the past experience, it is known that
the probabilities that he will come by train, bus, scooter or by other means of transport
are respectively 3 1 1
2
, ,
10 5 10and
5 |
1 | 7168-7171 | 0140
28
0 0345
69
=
Example 20 A doctor is to visit a patient From the past experience, it is known that
the probabilities that he will come by train, bus, scooter or by other means of transport
are respectively 3 1 1
2
, ,
10 5 10and
5 The probabilities that he will be late are 1 1
, , and1
4 3
12,
if he comes by train, bus and scooter respectively, but if he comes by other means of
transport, then he will not be late |
1 | 7169-7172 | 0345
69
=
Example 20 A doctor is to visit a patient From the past experience, it is known that
the probabilities that he will come by train, bus, scooter or by other means of transport
are respectively 3 1 1
2
, ,
10 5 10and
5 The probabilities that he will be late are 1 1
, , and1
4 3
12,
if he comes by train, bus and scooter respectively, but if he comes by other means of
transport, then he will not be late When he arrives, he is late |
1 | 7170-7173 | From the past experience, it is known that
the probabilities that he will come by train, bus, scooter or by other means of transport
are respectively 3 1 1
2
, ,
10 5 10and
5 The probabilities that he will be late are 1 1
, , and1
4 3
12,
if he comes by train, bus and scooter respectively, but if he comes by other means of
transport, then he will not be late When he arrives, he is late What is the probability
that he comes by train |
1 | 7171-7174 | The probabilities that he will be late are 1 1
, , and1
4 3
12,
if he comes by train, bus and scooter respectively, but if he comes by other means of
transport, then he will not be late When he arrives, he is late What is the probability
that he comes by train Solution Let E be the event that the doctor visits the patient late and let T1, T2, T3, T4
be the events that the doctor comes by train, bus, scooter, and other means of transport
respectively |
1 | 7172-7175 | When he arrives, he is late What is the probability
that he comes by train Solution Let E be the event that the doctor visits the patient late and let T1, T2, T3, T4
be the events that the doctor comes by train, bus, scooter, and other means of transport
respectively Then
P(T1) =
2
3
4
3
1
1
2
, P(T )
,P(T )
and P(T )
10
5
10
5
=
=
=
(given)
P(E|T1) = Probability that the doctor arriving late comes by train = 1
4
Similarly, P(E|T2) = 1
3 , P(E|T3) = 1
12 and P(E|T4) = 0, since he is not late if he
comes by other means of transport |
1 | 7173-7176 | What is the probability
that he comes by train Solution Let E be the event that the doctor visits the patient late and let T1, T2, T3, T4
be the events that the doctor comes by train, bus, scooter, and other means of transport
respectively Then
P(T1) =
2
3
4
3
1
1
2
, P(T )
,P(T )
and P(T )
10
5
10
5
=
=
=
(given)
P(E|T1) = Probability that the doctor arriving late comes by train = 1
4
Similarly, P(E|T2) = 1
3 , P(E|T3) = 1
12 and P(E|T4) = 0, since he is not late if he
comes by other means of transport Therefore, by Bayes' Theorem, we have
P(T1|E) = Probability that the doctor arriving late comes by train
=
1
1
1
1
2
2
3
3
4
4
P(T )P(E|T )
P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T )
=
3
1
10
4
3
1
1
1
1
1
2
0
10
4
5
3
10
12
5
= 3
120
1
40
18
2
×
=
Hence, the required probability is 1
2 |
1 | 7174-7177 | Solution Let E be the event that the doctor visits the patient late and let T1, T2, T3, T4
be the events that the doctor comes by train, bus, scooter, and other means of transport
respectively Then
P(T1) =
2
3
4
3
1
1
2
, P(T )
,P(T )
and P(T )
10
5
10
5
=
=
=
(given)
P(E|T1) = Probability that the doctor arriving late comes by train = 1
4
Similarly, P(E|T2) = 1
3 , P(E|T3) = 1
12 and P(E|T4) = 0, since he is not late if he
comes by other means of transport Therefore, by Bayes' Theorem, we have
P(T1|E) = Probability that the doctor arriving late comes by train
=
1
1
1
1
2
2
3
3
4
4
P(T )P(E|T )
P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T )
=
3
1
10
4
3
1
1
1
1
1
2
0
10
4
5
3
10
12
5
= 3
120
1
40
18
2
×
=
Hence, the required probability is 1
2 © NCERT
not to be republished
PROBABILITY 555
Example 21 A man is known to speak truth 3 out of 4 times |
1 | 7175-7178 | Then
P(T1) =
2
3
4
3
1
1
2
, P(T )
,P(T )
and P(T )
10
5
10
5
=
=
=
(given)
P(E|T1) = Probability that the doctor arriving late comes by train = 1
4
Similarly, P(E|T2) = 1
3 , P(E|T3) = 1
12 and P(E|T4) = 0, since he is not late if he
comes by other means of transport Therefore, by Bayes' Theorem, we have
P(T1|E) = Probability that the doctor arriving late comes by train
=
1
1
1
1
2
2
3
3
4
4
P(T )P(E|T )
P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T )
=
3
1
10
4
3
1
1
1
1
1
2
0
10
4
5
3
10
12
5
= 3
120
1
40
18
2
×
=
Hence, the required probability is 1
2 © NCERT
not to be republished
PROBABILITY 555
Example 21 A man is known to speak truth 3 out of 4 times He throws a die and
reports that it is a six |
1 | 7176-7179 | Therefore, by Bayes' Theorem, we have
P(T1|E) = Probability that the doctor arriving late comes by train
=
1
1
1
1
2
2
3
3
4
4
P(T )P(E|T )
P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T )+P(T )P(E|T )
=
3
1
10
4
3
1
1
1
1
1
2
0
10
4
5
3
10
12
5
= 3
120
1
40
18
2
×
=
Hence, the required probability is 1
2 © NCERT
not to be republished
PROBABILITY 555
Example 21 A man is known to speak truth 3 out of 4 times He throws a die and
reports that it is a six Find the probability that it is actually a six |
1 | 7177-7180 | © NCERT
not to be republished
PROBABILITY 555
Example 21 A man is known to speak truth 3 out of 4 times He throws a die and
reports that it is a six Find the probability that it is actually a six Solution Let E be the event that the man reports that six occurs in the throwing of the
die and let S1 be the event that six occurs and S2 be the event that six does not occur |
1 | 7178-7181 | He throws a die and
reports that it is a six Find the probability that it is actually a six Solution Let E be the event that the man reports that six occurs in the throwing of the
die and let S1 be the event that six occurs and S2 be the event that six does not occur Then
P(S1) = Probability that six occurs = 1
6
P(S2) = Probability that six does not occur = 5
6
P(E|S1) = Probability that the man reports that six occurs when six has
actually occurred on the die
= Probability that the man speaks the truth = 3
4
P(E|S2) = Probability that the man reports that six occurs when six has
not actually occurred on the die
= Probability that the man does not speak the truth
3
1
1
4
4
Thus, by Bayes' theorem, we get
P(S1|E) = Probability that the report of the man that six has occurred is
actually a six
=
1
1
1
1
2
2
P(S )P(E |S )
P(S )P(E|S )+P(S )P(E|S )
=
1
3
1
24
3
6
4
1
3
5
1
8
8
8
6
4
6
4
Hence, the required probability is 3 |
1 | 7179-7182 | Find the probability that it is actually a six Solution Let E be the event that the man reports that six occurs in the throwing of the
die and let S1 be the event that six occurs and S2 be the event that six does not occur Then
P(S1) = Probability that six occurs = 1
6
P(S2) = Probability that six does not occur = 5
6
P(E|S1) = Probability that the man reports that six occurs when six has
actually occurred on the die
= Probability that the man speaks the truth = 3
4
P(E|S2) = Probability that the man reports that six occurs when six has
not actually occurred on the die
= Probability that the man does not speak the truth
3
1
1
4
4
Thus, by Bayes' theorem, we get
P(S1|E) = Probability that the report of the man that six has occurred is
actually a six
=
1
1
1
1
2
2
P(S )P(E |S )
P(S )P(E|S )+P(S )P(E|S )
=
1
3
1
24
3
6
4
1
3
5
1
8
8
8
6
4
6
4
Hence, the required probability is 3 8
EXERCISE 13 |
1 | 7180-7183 | Solution Let E be the event that the man reports that six occurs in the throwing of the
die and let S1 be the event that six occurs and S2 be the event that six does not occur Then
P(S1) = Probability that six occurs = 1
6
P(S2) = Probability that six does not occur = 5
6
P(E|S1) = Probability that the man reports that six occurs when six has
actually occurred on the die
= Probability that the man speaks the truth = 3
4
P(E|S2) = Probability that the man reports that six occurs when six has
not actually occurred on the die
= Probability that the man does not speak the truth
3
1
1
4
4
Thus, by Bayes' theorem, we get
P(S1|E) = Probability that the report of the man that six has occurred is
actually a six
=
1
1
1
1
2
2
P(S )P(E |S )
P(S )P(E|S )+P(S )P(E|S )
=
1
3
1
24
3
6
4
1
3
5
1
8
8
8
6
4
6
4
Hence, the required probability is 3 8
EXERCISE 13 3
1 |
1 | 7181-7184 | Then
P(S1) = Probability that six occurs = 1
6
P(S2) = Probability that six does not occur = 5
6
P(E|S1) = Probability that the man reports that six occurs when six has
actually occurred on the die
= Probability that the man speaks the truth = 3
4
P(E|S2) = Probability that the man reports that six occurs when six has
not actually occurred on the die
= Probability that the man does not speak the truth
3
1
1
4
4
Thus, by Bayes' theorem, we get
P(S1|E) = Probability that the report of the man that six has occurred is
actually a six
=
1
1
1
1
2
2
P(S )P(E |S )
P(S )P(E|S )+P(S )P(E|S )
=
1
3
1
24
3
6
4
1
3
5
1
8
8
8
6
4
6
4
Hence, the required probability is 3 8
EXERCISE 13 3
1 An urn contains 5 red and 5 black balls |
1 | 7182-7185 | 8
EXERCISE 13 3
1 An urn contains 5 red and 5 black balls A ball is drawn at random, its colour is
noted and is returned to the urn |
1 | 7183-7186 | 3
1 An urn contains 5 red and 5 black balls A ball is drawn at random, its colour is
noted and is returned to the urn Moreover, 2 additional balls of the colour drawn
are put in the urn and then a ball is drawn at random |
1 | 7184-7187 | An urn contains 5 red and 5 black balls A ball is drawn at random, its colour is
noted and is returned to the urn Moreover, 2 additional balls of the colour drawn
are put in the urn and then a ball is drawn at random What is the probability that
the second ball is red |
1 | 7185-7188 | A ball is drawn at random, its colour is
noted and is returned to the urn Moreover, 2 additional balls of the colour drawn
are put in the urn and then a ball is drawn at random What is the probability that
the second ball is red © NCERT
not to be republished
556
MATHEMATICS
2 |
1 | 7186-7189 | Moreover, 2 additional balls of the colour drawn
are put in the urn and then a ball is drawn at random What is the probability that
the second ball is red © NCERT
not to be republished
556
MATHEMATICS
2 A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black
balls |
1 | 7187-7190 | What is the probability that
the second ball is red © NCERT
not to be republished
556
MATHEMATICS
2 A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black
balls One of the two bags is selected at random and a ball is drawn from the bag
which is found to be red |
1 | 7188-7191 | © NCERT
not to be republished
556
MATHEMATICS
2 A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black
balls One of the two bags is selected at random and a ball is drawn from the bag
which is found to be red Find the probability that the ball is drawn from the
first bag |
1 | 7189-7192 | A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black
balls One of the two bags is selected at random and a ball is drawn from the bag
which is found to be red Find the probability that the ball is drawn from the
first bag 3 |
1 | 7190-7193 | One of the two bags is selected at random and a ball is drawn from the bag
which is found to be red Find the probability that the ball is drawn from the
first bag 3 Of the students in a college, it is known that 60% reside in hostel and 40% are
day scholars (not residing in hostel) |
1 | 7191-7194 | Find the probability that the ball is drawn from the
first bag 3 Of the students in a college, it is known that 60% reside in hostel and 40% are
day scholars (not residing in hostel) Previous year results report that 30% of all
students who reside in hostel attain A grade and 20% of day scholars attain A
grade in their annual examination |
1 | 7192-7195 | 3 Of the students in a college, it is known that 60% reside in hostel and 40% are
day scholars (not residing in hostel) Previous year results report that 30% of all
students who reside in hostel attain A grade and 20% of day scholars attain A
grade in their annual examination At the end of the year, one student is chosen
at random from the college and he has an A grade, what is the probability that the
student is a hostlier |
1 | 7193-7196 | Of the students in a college, it is known that 60% reside in hostel and 40% are
day scholars (not residing in hostel) Previous year results report that 30% of all
students who reside in hostel attain A grade and 20% of day scholars attain A
grade in their annual examination At the end of the year, one student is chosen
at random from the college and he has an A grade, what is the probability that the
student is a hostlier 4 |
1 | 7194-7197 | Previous year results report that 30% of all
students who reside in hostel attain A grade and 20% of day scholars attain A
grade in their annual examination At the end of the year, one student is chosen
at random from the college and he has an A grade, what is the probability that the
student is a hostlier 4 In answering a question on a multiple choice test, a student either knows the
answer or guesses |
1 | 7195-7198 | At the end of the year, one student is chosen
at random from the college and he has an A grade, what is the probability that the
student is a hostlier 4 In answering a question on a multiple choice test, a student either knows the
answer or guesses Let 3
4 be the probability that he knows the answer and 1
4
be the probability that he guesses |
1 | 7196-7199 | 4 In answering a question on a multiple choice test, a student either knows the
answer or guesses Let 3
4 be the probability that he knows the answer and 1
4
be the probability that he guesses Assuming that a student who guesses at the
answer will be correct with probability 1
4 |
1 | 7197-7200 | In answering a question on a multiple choice test, a student either knows the
answer or guesses Let 3
4 be the probability that he knows the answer and 1
4
be the probability that he guesses Assuming that a student who guesses at the
answer will be correct with probability 1
4 What is the probability that the stu-
dent knows the answer given that he answered it correctly |
1 | 7198-7201 | Let 3
4 be the probability that he knows the answer and 1
4
be the probability that he guesses Assuming that a student who guesses at the
answer will be correct with probability 1
4 What is the probability that the stu-
dent knows the answer given that he answered it correctly 5 |
1 | 7199-7202 | Assuming that a student who guesses at the
answer will be correct with probability 1
4 What is the probability that the stu-
dent knows the answer given that he answered it correctly 5 A laboratory blood test is 99% effective in detecting a certain disease when it is
in fact, present |
1 | 7200-7203 | What is the probability that the stu-
dent knows the answer given that he answered it correctly 5 A laboratory blood test is 99% effective in detecting a certain disease when it is
in fact, present However, the test also yields a false positive result for 0 |
1 | 7201-7204 | 5 A laboratory blood test is 99% effective in detecting a certain disease when it is
in fact, present However, the test also yields a false positive result for 0 5% of
the healthy person tested (i |
1 | 7202-7205 | A laboratory blood test is 99% effective in detecting a certain disease when it is
in fact, present However, the test also yields a false positive result for 0 5% of
the healthy person tested (i e |
1 | 7203-7206 | However, the test also yields a false positive result for 0 5% of
the healthy person tested (i e if a healthy person is tested, then, with probability
0 |
1 | 7204-7207 | 5% of
the healthy person tested (i e if a healthy person is tested, then, with probability
0 005, the test will imply he has the disease) |
1 | 7205-7208 | e if a healthy person is tested, then, with probability
0 005, the test will imply he has the disease) If 0 |
1 | 7206-7209 | if a healthy person is tested, then, with probability
0 005, the test will imply he has the disease) If 0 1 percent of the population
actually has the disease, what is the probability that a person has the disease
given that his test result is positive |
1 | 7207-7210 | 005, the test will imply he has the disease) If 0 1 percent of the population
actually has the disease, what is the probability that a person has the disease
given that his test result is positive 6 |
1 | 7208-7211 | If 0 1 percent of the population
actually has the disease, what is the probability that a person has the disease
given that his test result is positive 6 There are three coins |
1 | 7209-7212 | 1 percent of the population
actually has the disease, what is the probability that a person has the disease
given that his test result is positive 6 There are three coins One is a two headed coin (having head on both faces),
another is a biased coin that comes up heads 75% of the time and third is an
unbiased coin |
1 | 7210-7213 | 6 There are three coins One is a two headed coin (having head on both faces),
another is a biased coin that comes up heads 75% of the time and third is an
unbiased coin One of the three coins is chosen at random and tossed, it shows
heads, what is the probability that it was the two headed coin |
1 | 7211-7214 | There are three coins One is a two headed coin (having head on both faces),
another is a biased coin that comes up heads 75% of the time and third is an
unbiased coin One of the three coins is chosen at random and tossed, it shows
heads, what is the probability that it was the two headed coin 7 |
1 | 7212-7215 | One is a two headed coin (having head on both faces),
another is a biased coin that comes up heads 75% of the time and third is an
unbiased coin One of the three coins is chosen at random and tossed, it shows
heads, what is the probability that it was the two headed coin 7 An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000
truck drivers |
1 | 7213-7216 | One of the three coins is chosen at random and tossed, it shows
heads, what is the probability that it was the two headed coin 7 An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000
truck drivers The probability of an accidents are 0 |
1 | 7214-7217 | 7 An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000
truck drivers The probability of an accidents are 0 01, 0 |
1 | 7215-7218 | An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000
truck drivers The probability of an accidents are 0 01, 0 03 and 0 |
1 | 7216-7219 | The probability of an accidents are 0 01, 0 03 and 0 15 respectively |
1 | 7217-7220 | 01, 0 03 and 0 15 respectively One of the insured persons meets with an accident |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.