Question Set B

cbse 12th Standard Maths Probability Important Questions And Answers Study Material - QB365

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Questions:

3 Marks

Bag A contains 6 red and 5 blue balls and another bag B contains 5 red and 8 blue balls. A ball id drawn from bag A without seeing its color and it is put into the bag B. Then a ball is drawn from bag B at random. Find the probability that the ball drawn is blue in colour.

There are 2000 scooter drivers, 4000 car drivers and 6000 truck drivers all insured. The probabilities of an accident involving a scooter, a car, a truck are 0.01, 0.03, 0.15 respectively. One of the insured drivers meets with an accident. What is the probability that he is a scooter driver?

In a bulb factory machines A, B and C manufacture 60%, 30% and 10% bulbs respectively. 1%, 2% and 3% of the bulbs produced respectively by A, B and C are found to be defective. Find the probability that this bulb was produced by the machine A.

Three bags contain balls as shown in the table below.

Bag	Number of white balls	Number of black balls	Number of Red balls
I	1	2	3
II	2	1	1
III	4	3	2

A bag is chosen art random and two balls are drawn from it. They happen to be white and red. What is the probability that they came from the III bag?

In a group of 400 people, 160 are smokers and non-vegetarian, 100 are smokers and vegetarian and the remaining are non-smokers and vegetarian. The probabilities of getting a special chest disease are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian? What value is reflected in this question?

Two cards are drawn from a well-shuffled pack of 52 cards one after the other without replacement. Find the probability that one of these is a queen and the other is a king of opposite color.

A doctor is to visit a patient. Form 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 \({3\over10},{1\over5},{1\over10} and {2\over5}\). The probabilities that he will be late are \({1\over4},{1\over3},{1\over12}\) 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?

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 the people free of HIV, 99% of the tests are judged HIV -ve but 1% are diagnosed as showing HIV +ve. 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 +ve. What is the probability that the person actually has HIV?

A letter is known to have come either from TATANAGAR or from CALCUTTA. On the envelope just two consecutive letters TA are visible. What is the probability that the letters came from TATANAGAR?

10.

There are two bags, one of which contains 3 black and 4 white balls, while the other contains 4 black and 3 white balls. Adie is thrown. It shows up 1or 3, a ball is taken from the first bag, but it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.

2 Marks

11.

Given P(A) = \(1\over2\), P(B) = \(1\over3\) and \(P(A\cap B)={1\over6}\) Are the events A and B independent?

12.

Given P(A) = 0.4, P(B) = 0.7 and P(B/A) = 0.6, Find \(P(A\cup B)\)

13.

A couple has 2 children. Find the probability that both are boys, if it is known that (a) one of them is a boy (b) the older child is boys.

14.

Two cards are drawn at random from a pack of 52 cards one-by-one without replacement. What is the probability of getting first card red and second card jack?

15.

A speaks truth in 80% cases and B speaks truth in 90% cases, In what percentage of cases are they likely to agree with each other in stating the same fact?

5 Marks

16.

A bag I contains 5 red and 4 white balls and a bag II contains 3 red and 3 white balls. Two balls are transferred from the bag I to the bag II and then one ball is drawn from bag II. If the ball drawn from the bag II is red, then find the probability that one red ball and one white ball are transferred from the bag I to the bag II.

17.

There are three coins. First is a biased that comes up tails 60% of the times, second is also a biased coin that comes up heads 75% of the times 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 first coin?

18.

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 two balls are drawn at random (without replacement) from the bag which are both found to be red. Find the probability that the balls are drawn from the first bag.

19.

A shopkeeper sells three types of flower seeds A₁, A₂ and A₃. They are sold as a mixture, where the proportions are 4 : 4 : 2, respectively. The germination rates of the three types of seedsare 45%, 60% and 35%. Calculate the probability.
(i) of a randomly chosen seed to germinate
(ii) that it willhot germinate given that the seed is of type A_3.
(iii) that it is of the type A₂ given that a randomly chosen seed does not germinate.

20.

Among 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 exams. At the end of year, one student is chosen at random from the college and he has A grade, what is the probability that the student is a hosteler?

Case Study Questions

21.

A pharmaceutical company wants to advertise a new product on T.V, where the product is specially designed for women. For that an advertising executive is hired to study television-viewing habits of married couples during prime time hours. Based on past viewing records he has determined that during prime time husbands are watching television 70% of the time. It has also been determined that when the husband is watching television, 30% of the time the wife is also watching. When the husband is not watching television, 40% of the time the wife is watching television.

Based on the above information, answer the following questions.
(i) The probability that the husband is not watching television during prime time, is

(a) 0.6

(b) 0.3

(c) 0.4

(d) 0.5

(ii) If the wife is watching television, the probability that husband is also watching television, is

(a) 2/11

(b) 7/11

(c) 5/11

(d) 8/11

(iii) The probability that both husband and wife are watching television during prime time, is

(a) 0.21

(b) 0.5

(c) 0.3

(d) 0.4

(iv) The probability that the wife is watching television during prime time, is

(a) 0.24

(b) 0.33

(c) 0.3

(d) 0.4

(v) If the wife is watching television, then the probability that husband is not watching television, is

(a) 2/11

(b) 4/11

(c) 1/11

(d) 5/11

22.

In a bilateral cricket series between India and South Africa, the probability that India wins the first match is 0.6. If India wins any match, then the probability that it wins the next match is 0.4, otherwise the probability is 0.3. Aso, it is given that there is no tie in any match

Based on the above information answer the following questions.
(i) The probability that India won the second match, if India has already loose the first match is

(a) 0.5

(b) 0.3

(c) 0.4

(d) 0.7

(ii) The probability that India losing the third match, if India has already loose the first two matches is

(a) 0.2

(b) 0.3

(c) 0.4

(d) 0.7

(iii) The probability th-.at India losing the first two matches is

(a) 0.12

(b) 0.28

(c) 0.42

(d) 0.01

(iv) The probability that India winning the first three matches is

(a) 0.92

(b) 0.96

(c) 0.94

(d) 0.096

(v) The probability that India winning exactly one of the first three matches is

(a) 0.205

(b) 0.21

(c) 0.408

(d) 0.312

23.

Box I contains 1 white, 3 black and 2 red balls. Box II contains 2 white, 1 black and 3 red balls. Box III contains 3 white, 2 black and 1 red balls. One box is chosen at random and two balls are drawn with replacement.

If E₁,E₂, E₃ be the events that the balls drawn from box 1,box II and box III respectively and E be the event that balls drawn are one white and one red, then answer the following questions.
(i) Probability of occurrence of everit E given that the balls drawn are from box I, is

(a) \(\frac{1}{9}\)

(b) \(\frac{2}{6}\)

(d) \(\frac{1}{7}\)

(ii) Probability of occurrence of event E, given that the balls drawn are from box II, is

(a) \(\frac{1}{3}\)

(b) \(\frac{1}{4}\)

(d) \(\frac{3}{5}\)

(iii) Probability of occurrence of event E, given that balls drawn are from box III, is

(a) \(\frac{1}{12}\)

(b) \(\frac{3}{11}\)

(d) \(\frac{4}{11}\)

(iv) The value of \(\sum_{i=1}^{3} P\left(E \mid E_{i}\right)\) is equal to

(a) \(\frac{5}{18}\)

(b) \(\frac{1}{2}\)

(d) \(\frac{11}{18}\)

(v) The probability that the balls drawn are from box II, given that event E has already occurred, is

(a) \(\frac{1}{11}\)

(b) \(\frac{6}{11}\)

(d) None of these

24.

A building contractor undertakes a job to construct 4 flats on a plot along with parking area, Due to strike, the probability of many construction workers not being present for the job is 0.65.
The probability that many are not present and still the work gets completed on time is 0.35. The probability that work will be completed on time when all workers are present is 0.80.
Let E₁ : represents the event when many workers were not present for the job;
E₂ : represents the event when all workers were present; and
E : represents completing the construction work on time.
Based on the above information, answer the following questions:
(i) What is the probability that all the workers are present for the job?
(ii) What is the probability that construction will be completed on time?
(iii) (a) What is the probability that many workers are not present given that the construction work is completed on time?
Or (b) What is the probability that all workers were present given that the construction job was completed on time?

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Answers Key:

3 Marks

Bag A. : 6 red + 5 blue; Bag B : 5 red + 8 blue
let a red ball be drawn from bag A
\(P\left(\mathrm{red}_{A}\right)=\frac{6}{11}\)
Number of balls in bag B: (5+1) red + 8 blue
\(P\left(\text { blue }_{B}\right)=\frac{8}{14}\)
therefore probability of drawing a blue ball from bag B, when a red ball is transferred from bag A to bag B is
\(=P\left(\text { red }_{A}\right) \cdot P\left(\text { blue }_{B}\right) \)
\(=\frac{6}{11} \times \frac{8}{14} \)
let blue ball is drawn from bag A
\(P\left(\text { blue }_{A}\right)=\frac{5}{11}\)
Number of balls in bag B : 5 red + (8 + I) blue
\(P\left(\text { blue }_{B}\right)=\frac{9}{14}\)
Probability of drawing a blue ball from bag B when a blue ball is transferred from bag A to bag B is
\(=P\left(\text { blue }_{A}\right) \cdot P\left(\text { blue }_{B}\right)\)
\(=\frac{5}{11} \times \frac{9}{14} \)
Hence, probability of drawing a blue ball from bag B when a ball is transferred from bag A to bag B = \(\frac{6}{11} \times \frac{8}{14}+\frac{5}{11} \times \frac{9}{14}=\frac{48+45}{154}=\frac{93}{154}\)

E : accident; S : scooter driver; C : car driver; T: truck driver
\(P(S)=\frac{2000}{12000}=\frac{2}{12} ; P(C)=\frac{4000}{12000}=\frac{4}{12} ; \)
\(P(T)=\frac{6000}{12000}=\frac{6}{12}\)
P(EIS) = 0.01; P(EIC) = 0.03; P(EIT) = 0.15
Using Bayes' Theorem the probability of accident of a scooter driver is
\(P(S / E) =\frac{P(S) \cdot P(E / S)}{P(S) \cdot P(E / S)+P(C) \cdot P(E / C)+P(T) \cdot P(E / T)} \)
\(=\frac{\frac{2}{12} \times 0.01}{\frac{2}{12} \times 0.01+\frac{4}{12} \times 0.03+\frac{6}{12} \times 0.15} \)
\(=\frac{0.02}{0.02+0.12+0.90}=\frac{2}{104}=\frac{1}{52} \)

\(\begin{array}{|c|c|c|c|} \hline \text { Machine } & \boldsymbol{A} & \boldsymbol{B} & \boldsymbol{C} \\ \hline \text { Production } & 60 \% & 30 \% & 10 \% \\ \hline \text { Defective } & 1 \% & 2 \% & 3 \% \\ \hline \end{array}\)
\(P(A)=\frac{60}{100}=\frac{6}{10}, P(B)=\frac{30}{100}=\frac{3}{10}, P(C)=\frac{10}{100}=\frac{1}{10}\)
E: bulb is defective
\(P(E / A)=\frac{1}{100}, P(E / B)=\frac{2}{100}, P(E / C)=\frac{3}{100}\)
Using Bayes' Theorem, probability that defective bulbs was produced by machine A,
\(\begin{array}{r} P(A / E)=\frac{P(A) \cdot P(E / A)}{P(A) \cdot P(E / A)+P(B) \cdot P(E / B)} \\ +P(C) \cdot P(E / C) \end{array}\)
\(=\frac{\frac{6}{10} \times \frac{1}{100}}{\frac{6}{10} \times \frac{1}{100}+\frac{3}{10} \times \frac{2}{100}+\frac{1}{10} \times \frac{3}{100}}=\frac{2}{5}\)

Bag	Number of white balls	Number of black balls	Number of Red balls
I	1	2	3
II	2	1	1
III	4	3	2

Probability of chosing a bag:
\(P(I)=\frac{1}{3}, P(I I)=\frac{1}{3}, P(I I I)=\frac{1}{3} .\)
E : one white and one red ball is drawn.
\(P(E / I)=\frac{{ }^{1} C_{1} \times{ }^{3} C_{1}}{{ }^{6} C_{2}}=\frac{1 \times 3 \times 2}{6 \times 5}=\frac{1}{5}\)
\(P(E / I I)=\frac{{ }^{2} C_{1} \times{ }^{1} C_{1}}{{ }^{4} C_{2}}=\frac{2 \times 1}{6}=\frac{1}{3} \)
\(P(E / I I)=\frac{{ }^{4} C_{1} \times{ }^{2} C_{1}}{{ }^{9} C_{2}}=\frac{4 \times 2 \times 2}{9 \times 8}=\frac{2}{9} \)
Using Bayes' Theorem,
Probability of drawing one white and one red from bag III is
\(P(I I I E) =\frac{P(I I I) \cdot P(E / I I I)}{P(I) \cdot P(E / I)+P(I I) \cdot P(E / I I)+P(I I I) \cdot P(E / I I I)} \)
\(=\frac{\frac{1}{3} \times \frac{2}{9}}{\frac{1}{3} \times \frac{1}{5}+\frac{1}{3} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{9}}=\frac{\frac{2}{9}}{\frac{1}{5}+\frac{1}{3}+\frac{2}{9}} \)
\(=\frac{\frac{2}{9}}{\frac{9+15+10}{45}}=\frac{2}{9} \times \frac{45}{34}=\frac{5}{17} \)

\(28\over45\)

P(a king and other a queen of opposite colour or a queen and other a king of opposite colour)
\(=\frac{4}{52} \times \frac{2}{51}+\frac{4}{52} \times \frac{2}{51} \)
\(=\frac{4}{663} \)

EI : train. Ec : bus, E₃ : scooter,
E.j : other means; E : arrives late
\(P\left(E_{1}\right)=\frac{3}{10} ; P\left(E_{2}\right)=\frac{1}{5} ; P\left(E_{3}\right)=\frac{1}{10} ; P\left(E_{4}\right)=\frac{2}{5} ; \)
\(P\left(E / E_{1}\right)=\frac{1}{4} ; P\left(E / E_{2}\right)=\frac{1}{3} ; P\left(E / E_{3}\right)=\frac{1}{12} ; P\left(E / E_{4}\right)=0 \)
Using Bayes' Theorem, probability that doctor is late and come by train
\(\begin{aligned} P\left(E_{1} / E\right)=& \frac{P\left(E_{1}\right) \cdot P\left(E / E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(E / E_{1}\right)+P\left(E_{2}\right) \cdot P\left(E / E_{2}\right)} \\ &+P\left(E_{3}\right) \cdot P\left(E / E_{3}\right)+P\left(E_{4}\right) \cdot P\left(E / E_{4}\right) \end{aligned}\)
\(=\frac{\frac{3}{10} \times \frac{1}{4}}{\frac{3}{10} \times \frac{1}{4}+\frac{1}{5} \times \frac{1}{3}+\frac{1}{10} \times \frac{1}{12}+\frac{2}{5} \times 0}=\frac{1}{2}\)

A: people having HIV
\(P(A)=\frac{0.1}{100}=\frac{1}{1000}\)
B: people free from HIV
\(P(B)=\frac{999}{1000}\)
E: test is HIV positive
\(P(E / A)=\frac{90}{100} ; P(E / B)=\frac{1}{100}\)
Using Bayes' Theorem, probability of reporting HIV positive when he actually has it.
\(P(A / E)=\frac{\frac{1}{1000} \times \frac{90}{100}}{\frac{1}{1000} \times \frac{90}{100}+\frac{999}{1000} \times \frac{1}{100}}=\frac{90}{1089}=\frac{10}{121}\)

Let E_I = Letter has come from CALCUITA
E₂ = Letter has come from TATANAGAR
and E = Two consecutive letters (i.e. alphabets) TA are visible on envelope
\(\therefore P\left(E_{1}\right)=\frac{1}{2}, P\left(E_{2}\right)=\frac{1}{2}, P\left(\frac{E}{E_{1}}\right)=\frac{n\left(E \cap E_{1}\right)}{n\left(E_{1}\right)}=\frac{1}{7}\)
[\(\therefore\) pairs of consecutive letters are CA, AL, LC, CU, UT, TT, TA]
and \(P\left(\frac{E}{E_{2}}\right)=\frac{n\left(E \cap E_{2}\right)}{n\left(E_{2}\right)}=\frac{2}{8}\)
[\(\therefore\)8 pairs of consecutive letters are TA, AT, TA;AN, NA, AG, GA, AR]
\(\therefore P\left(\frac{E_{2}}{E}\right)=\frac{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}\)
[using Baye's theorem]
\(=\frac{\frac{1}{2} \times \frac{2}{8}}{\frac{1}{2} \times \frac{1}{7}+\frac{1}{2} \times \frac{2}{8}}=\frac{\frac{2}{16}}{\frac{8+14}{2 \times 7 \times 8}}=\frac{2 \times 7}{22}=\frac{7}{11}\)
Hence, the probability that the letter TA came from TATANAGAR is \(\frac{7}{11}\).

10.

Let E₁ = Die shows up 1 or 3
E₂ = Die shows up 2, 4, 5, 6
and A = Black ball is drawn
Then, \(P\left(E_{1}\right)=\frac{2}{6}, P\left(E_{2}\right)=\frac{4}{6}, P\left(\frac{A}{E_{1}}\right)=\frac{3}{7}, P\left(\frac{A}{E_{2}}\right)=\frac{4}{7}\)
\(\therefore\) Required probability
\(P(A)=P\left(E_{1}\right) \cdot P\left(\frac{A}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{A}{E_{2}}\right)\)
= \(\frac{11}{12}\)

2 Marks

11.

P(A) ⋅ P(B) = \(1\over2\)⋅\(1\over3\) = \(1\over6\) = P(A∩B)
Yes, the events are independent.

12.

\( P(B / A)=\frac{P(A \cap B)}{P(A)} \)
\(\Rightarrow 0.6 \times 0.4=P(A \cap B) \)
\(\Rightarrow P(A \cap B)=0.24 \)
\( P(A \cup B)=P(A)+P(B)-P(A \cap B) \)
\(=0.4+0.7-0.24=0.86 \)

13.

Sample space ={B₁B₂, B₁G₂, G₁B₂, G₁G₂}, B₁ and G₁ are the older boy and girl respectively.
Let E₁ = both the children are boys;
E₂ = one of the children are boys;
E₃ = the older child is a boy
Then, (a) P(E₁/E₂) = \(P\left( \frac { { E }_{ 1 }\cap { E }_{ 2 } }{ { E }_{ 2 } } \right) =\frac { \frac { 1 }{ 4 } }{ \frac { 3 }{ 4 } } =\frac { 1 }{ 3 } \)
(b) P(E₁/E₃) = \(P\left( \frac { { E }_{ 1 }\cap { E }_{ 3 } }{ { E }_{ 3 } } \right) =\frac { \frac { 1 }{ 4 } }{ \frac { 2 }{ 4 } } =\frac { 1 }{ 2 } \)

14.

\(\because\) Two cards are drawn at random from a pack of 52 cards one-by-one without replacement.
\(\therefore\) The required probability = P {(the first is a red jack card and the second is a jack card) or (the first is a red non-jack card and the second is a jack card)}
\(=\frac{2}{52} \times \frac{3}{51}+\frac{24}{52} \times \frac{4}{51}=\frac{1}{26}\)

15.

Let A_T : Event that A speaks truth
and B_T : Event that B speaks truth.
Given, \(\begin{aligned} P\left(A_T\right)=\frac{80}{100}=\frac{4}{5} \end{aligned}\)
\(\begin{aligned} P\left(B_T\right)=\frac{90}{100}=\frac{9}{10} \end{aligned}\)
P(agree) = P(both speaking truth or both telling lie)
\(\begin{aligned} =P\left(A_T B_T \text { or } \bar{A}_T \bar{B}_T\right) \end{aligned}\)
\(\begin{aligned} =P\left(A_T\right) P\left(B_T\right) \text { or } P\left(\bar{A}_T\right) P\left(\bar{B}_T\right) \end{aligned}\)
\(\begin{aligned} =\left(\frac{4}{5}\right)\left(\frac{9}{10}\right)+\left(\frac{1}{5}\right)\left(\frac{1}{10}\right) \end{aligned}\)
\(\begin{aligned} =\frac{36+1}{50}=\frac{37}{50}=\frac{74}{100}=74 \% \end{aligned}\)

5 Marks

16.

Let, E₁: Two white balls are transferred
E₂: Two red balls are transferred
E₃: One red and one white ball are transferred.
A: The ball drawn from the bag II is red.
\(P({ E }_{ 1 })=\frac { { 4 }_{ C_{ 2 } } }{ { 9 }_{ { c }_{ 2 } } } =\frac { 4\times 3 }{ 9\times 8 } =\frac { 1 }{ 6 } \)
\({ P({ E } }_{ 2 })=\frac { { 5 }_{ C_{ 2 } } }{ { 9 }_{ C_{ 2 } } } =\frac { 4\times 3 }{ 9\times 8 } =\frac { 5 }{ 18 } \)
\(P(E_{ 3 })=\frac { { 5 }_{ { C }_{ 1 } }\times { 4 }_{ C_{ 1 } } }{ { 9 }_{ C_{ 2 } } } =\frac { 4\times 5\times 2 }{ 9\times 8 } =\frac { 5 }{ 9 } \)
\(P(A/{ E }_{ 1 })=\frac { 3 }{ 8 } ,P(A/{ E }_{ 2 })=\frac { 5 }{ 8 } ,\)
\(P(A/{ E }_{ 3 })=\frac { 4 }{ 8 } \)
The required probability, P(E₃/A), by Bayes' Theorem
\(=\frac { P({ E }_{ 3 }).P(A/{ E }_{ 3 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 })+P({ E }_{ 3 }).P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 5 }{ 9 } \times \frac { 4 }{ 8 } }{ \frac { 1 }{ 6 } \times \frac { 3 }{ 8 } +\frac { 5 }{ 18 } \times \frac { 5 }{ 8 } +\frac { 5 }{ 9 } \times \frac { 4 }{ 8 } } \)
\(=\frac { 20 }{ 37 } \)

17.

Let the events be:
E₁ = Choosing 1^st coin
E₂ = Choosing 2^nd coin
E₃ = Choosing 3^rd coin
A: Getting Heads
\(P({ E }_{ 1 })=P({ E }_{ 2 })=P(E_{ 3 })=\frac { 1 }{ 3 } \)
\(P(A/{ E }_{ 1 })=\frac { 40 }{ 100 } ,P(A/{ E }_{ 2 })=\frac { 75 }{ 100 } ,\)
\(P(A/{ E }_{ 3 })=\frac { 1 }{ 2 } \)
\(P({ E }_{ 1 }/A)\)
\(=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 })+P({ E }_{ 3 })P(A/{ E }_{ 3 }) } \)
\(=\frac { \frac { 1 }{ 3 } .\frac { 40 }{ 100 } }{ \frac { 1 }{ 3 } .\frac { 40 }{ 100 } +\frac { 1 }{ 3 } .\frac { 75 }{ 100 } +\frac { 1 }{ 3 } .\frac { 1 }{ 2 } } =\frac { 8 }{ 33 } \)

18.

Let E₁: Event selecting bag with 4 red & 4 black balls
E₂: Event selecting bag with 2 red & 6 black balls
A: Event selecting 2 red balls without replacement
Then, P(E₁) = P(E₂) = \(\frac { 1 }{ 2 } \)
\(P(A/{ E }_{ 1 })=\frac { 4_{ C_{ 2 } } }{ { 8 }_{ C_{ 2 } } } =\frac { 3 }{ 14 } ,\)
\(P(A/{ E }_{ 2 })=\frac { { 2 }_{ { C }_{ 2 } } }{ { 8 }_{ { C }_{ 2 } } } =\frac { 1 }{ 28 } .\)
\(P\left( \frac { { E }_{ 1 } }{ A } \right) =\frac { P({ E }_{ 1 }).P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 }) } \)
\(=\frac { \frac { 1 }{ 2 } .\frac { 3 }{ 14 } }{ \frac { 1 }{ 2 } .\frac { 3 }{ 14 } +\frac { 1 }{ 2 } .\frac { 1 }{ 28 } } =\frac { 6 }{ 7 } \)

19.

We have, A₁: A₂:A₃ = 4: 4: 2
\(\therefore P\left(A_{1}\right)=\frac{4}{10}, P\left(A_{2}\right)=\frac{4}{10} \text { and } P\left(A_{3}\right)=\frac{2}{10}\)
where A_I',A₂ and A₃ denote the event of choosing flower seeds A_1, A₂and A₃ respectively,
Let E be the event that a seed germinates and E be the event that a seed does not germinate.
Then,\(P\left(\frac{E}{A_{1}}\right)=\frac{45}{100}, P\left(\frac{E}{A_{2}}\right)=\frac{60}{100}, P\left(\frac{E}{A_{3}}\right)=\frac{35}{100}\)
and \(P\left(\frac{\bar{E}}{A_{1}}\right)=\frac{55}{100}, P\left(\frac{\bar{E}}{A_{2}}\right)=\frac{40}{100}, P\left(\frac{\bar{E}}{A_{3}}\right)=\frac{65}{100}\)
(i) Probability that a randomly chosen seed to germinate,
\(P(E)=P\left(A_{1}\right) \cdot\left(\frac{E}{A_{1}}\right)+P\left(A_{2}\right) \cdot P\left(\frac{E}{A_{2}}\right)+P\left(A_{3}\right) \cdot P\left(\frac{E}{A_{3}}\right)\)
\(=\frac{4}{10} \times \frac{45}{100}+\frac{4}{10} \times \frac{60}{100}+\frac{2}{10} \times \frac{35}{100}\)
\(=\frac{180}{1000}+\frac{240}{1000}+\frac{70}{1000}=\frac{490}{1000}=0.49\)
\(\text { (ii) } P\left(\frac{\bar{E}}{A_{3}}\right)=1-P\left(\frac{E}{A_{3}}\right)=1-\frac{35}{100}=\frac{65}{100}\)
\(\text { (iii) } P\left(\frac{A_{2}}{\bar{E}}\right)\)
\(=\frac{P\left(A_{2}\right) \cdot P\left(\frac{\bar{E}}{A_{2}}\right)}{P\left(A_{1}\right) \cdot P\left(\frac{\bar{E}}{A_{1}}\right)+P\left(A_{2}\right) \cdot P\left(\frac{\bar{E}}{A_{2}}\right)+P\left(A_{3}\right) \cdot P\left(\frac{\bar{E}}{A_{3}}\right)}\)
\(=\frac{\frac{4}{10} \times \frac{40}{100}}{\frac{4}{10} \times \frac{55}{100}+\frac{4}{10} \times \frac{40}{100}+\frac{2}{10} \times \frac{65}{100}}\)
\(=\frac{\frac{160}{1000}}{\frac{220}{1000}+\frac{160}{1000}+\frac{130}{1000}}=\frac{\frac{160}{1000}}{\frac{510}{1000}}=\frac{16}{51}\)
= 0.313725 = 0.314

20.

Let us detine the events as
E₁ : Students reside in a hostel
E₂ : Students are day scholars
A : Students get A grade
Then,
P(E₁) = Probability that student reside in a hostel
\(=60 \%=\frac{60}{100}\)
and P(E₂) = Probability that students are day scholars = 1 - \(\frac{60}{100}=\frac{40}{100}\)
Also, P(A/E₁) = Probability that hostelers get A grade
\(=30 \%=\frac{30}{100}\)
and P(A/E₂)= Probability that students having day scholars get A grade
\(=20 \%=\frac{20}{100}\)
\(\therefore\) The probability that the selecting student is a hosteler having A grade,
\(P\left(E_1 / A\right)=\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)}\)
[by Baye's theorem]
\(\begin{aligned} =\frac{\frac{60}{100} \times \frac{30}{100}}{\left(\frac{60}{100} \times \frac{30}{100}\right)+\left(\frac{40}{100} \times \frac{20}{100}\right)} \end{aligned}\)
\(\begin{aligned} =\frac{1800}{1800+800}=\frac{1800}{2600}=\frac{18}{26}=\frac{9}{13} \end{aligned}\)

Case Study Questions

21.

(i) (b): Since, it is given that during prime time husband is watching T.V. 70%of the time
∴ Required probability = 1 - P(husband is watching television during prime time)
= 1 - 0.7 = 0.3
(ii) (b): Let H be the event that husband is watching TV, W be the event that wife is watching TV.
Then, P(H) = 0.7, P(H) = 0.3
P(W I H) = 0.3 arltl P(W I H) = 0.4
∴ Required probability = P(H I W)
\(=\frac{P(\mathrm{H}) \cdot P(\mathrm{~W} \mid \mathrm{H})}{P(\mathrm{H}) \cdot P(\mathrm{~W} \mid \mathrm{H})+P(\overline{\mathrm{H}}) P(\mathrm{~W} \mid \overline{\mathrm{H}})} \)
\(=\frac{0.7 \times 0.3}{0.7 \times 0.3+0.4 \times 0.3}=\frac{0.21}{0.33}=\frac{7}{11}\)
(iii) (a) : Required probability = P(H \(\cap\) W)
= P(H)P(W I H) = 0.7 x 0.3 = 0.21
(iv) (b): Required probability = P(W) = \(\frac{P(\mathrm{H} \cap \mathrm{W})}{P(\mathrm{H} \mid \mathrm{W})}\)
\(=\frac{0.21}{7 / 11}=\frac{21}{100} \times \frac{11}{7}=\frac{33}{100}=0.33\)
(v) (b): Required probability = P(H I W)
\(=1-P(\mathrm{H} \mid \mathrm{W})=1-\frac{7}{11}=\frac{4}{11}\)

22.

(i) (c): It is given that if India loose any match, then the probability that it wins the next match is 0.3.
∴ Required probability = 0.3
(ii) (d): It is given that, if India loose any match, then the probability that it wins the next match is 0.3.
∴ Required probability = 1 - 0.3 = 0.7
(iii) (b): Required probability = P(lndia losing first match) . P(India losing second match when India has already lost first match)
= 0.4 x 0.7 = 0.28
(iv) (d): Required probability = P(lndia winning first match) . P(India winning second match if India has already won first match) P(lndia winning third match if India has already won first two matches)
= 0.6 x 0.4 x 0.4 = 0.096
(v) (c): Required probability = P(Win 1^st match) P(Lose 2^nd match) P(Lose 3^rd match) + P (Lose 1^st match) P(Win 2^nd match) P(Lose 3^rd match) + P(Lose 1^st match) P(Lose 2^nd match) P(Win 3^rd match)
= 0.6 x (1 - 0.4)・(1- 0.3) + (1 - 0.6)・(0.3)(1 - 0.4) + (1 - 0.6) (1 - 0.3) (0.3)
= 0.6 x 0.6 x 0.7 + 0.4 x 0.3 x 0.6 + 0.4 x 0.7 x 0.3
= 0.252 + 0.072 + 0.084 = 0.408

23.

We have, \(P\left(E_{1}\right)=P\left(E_{2}\right)=P\left(E_{3}\right)=\frac{1}{3}\)
(i) (a) : P(E l E₁) = Probability of drawing red and white ball, if box I is selected.
= P(red) x P(white) + P(white) x P(red)
\(=\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6}=\frac{4}{36}=\frac{1}{9}\)
(ii) (a): P(E l E₁) = Probability of drawing red and white balls, if box II is selected
= P(red) x P(white) + P(white) x P(red)
\(=\frac{3}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{3}{6}=\frac{12}{36}=\frac{1}{3}\)
(iii) (c): p(E IE3) = Probability of drawing red and white balls, if box III is selected.
= P(red) x P(white) + P(white) x P(red)
\(=\frac{1}{6} \times \frac{3}{6}+\frac{3}{6} \times \frac{1}{6}=\frac{6}{36}=\frac{1}{6}\)
(iv) (d): \(\sum_{i=1}^{3} P\left(E \mid E_{i}\right)=P\left(E \mid E_{1}\right)+P\left(E \mid E_{2}\right)+P\left(E \mid E_{3}\right)\)
\(=\frac{4}{36}+\frac{12}{36}+\frac{6}{36}=\frac{4+12+6}{36}=\frac{22}{36}=\frac{11}{18}\)
(v) (b): Using Bayes' theorem
\( P\left(E_{2} \mid E\right)=\frac{P\left(E \mid E_{2}\right) \cdot P\left(E_{2}\right)}{P\left(E \mid E_{1}\right) \cdot P\left(E_{1}\right)+P\left(E \mid E_{2}\right) \cdot P\left(E_{2}\right)+P\left(E \mid E_{3}\right) \cdot P\left(E_{3}\right)} \)
\(=\frac{\frac{1}{3} \times \frac{1}{3}}{\left(\frac{1}{9} \times \frac{1}{3}\right)+\left(\frac{1}{3} \times \frac{1}{3}\right)+\left(\frac{1}{6} \times \frac{1}{3}\right)} \)
\(=\frac{\frac{1}{3}}{\frac{1}{9}+\frac{1}{3}+\frac{1}{6}}=\frac{\frac{1}{3}}{\frac{11}{18}}=\frac{1}{3} \times \frac{18}{11}=\frac{6}{11}\)

24.

Given, P(E₁) = 0.65, P(E/E₁) = 0.35 and \(P\left(\frac{E}{E_2}\right)=0.80\)
(i) P(E₂) = 1 - P(E₁) = 1 - 0.65 = 0.35
(ii) \(\begin{aligned} P(E) & =P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) P\left(\frac{E}{E_2}\right) \end{aligned}\)
\(\begin{aligned} =0.65 \times 0.35+0.35 \times 0.80 \end{aligned}\)
\(\approx 0.23+0.28=0.51\)
(iii) (a) \(\begin{aligned} P\left(\frac{E_1}{E}\right) & =\frac{P\left(E_1\right) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) P\left(E / E_2\right)} \end{aligned}\)
\(\begin{aligned} =\frac{0.65 \times 0.35}{0.65 \times 0.35+0.35 \times 0.80}=0.45 \end{aligned}\)

or (b) \(\begin{aligned} P\left(\frac{E_2}{E}\right) & =\frac{P\left(E_2\right) P\left(E / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) P\left(E / E_2\right)} \end{aligned}\)
\(\begin{aligned} =\frac{0.35 \times 0.80}{0.65 \times 0.35+0.35 \times 0.80} \end{aligned}\)
= 0.55