Question 20:

If S is the sample space and P (A) = $\frac{1}{3}$ P (B) and S = A ∪ B, where A and B are two mutually exclusive events, then P (A) =
(a) 1/4
(b) 1/2
(c) 3/4
(d) 3/8

Answer:

$(a) \frac{1}{4} P (A) = \frac{1}{3} P (B) \Rightarrow P (B) = 3 P (A) . . . (1) A and B are mutually exclusive events. ⇒ P (A \cap B) = 0 Now, P (A \cup B) = P (A) + P (B) = P (S) \Rightarrow P (A) + P (B) = 1 \Rightarrow P (A) + 3 P (A) = 1 [From (1)] \Rightarrow 4 P (A) = 1 \Rightarrow P (A) = \frac{1}{4}$

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Question 21:

If A and B are two events, then P ( $A$ ∩ B) =
(a) P $(A)$ P $(B)$
(b) 1 − P (A) − P (B)
(c) P (A) + P (B) − P (A ∩ B)
(d) P (B) − P (A ∩ B)

Answer:

(d) P (B) − P (A ∩ B)

$From the diagram, we get A \cap B and \bar{A} \cap B are mutually exclusive events such that (A \cap B) \cup (\bar{A} \cap B) = B . Therefore by addition theorem of probability we have P (A \cap B) + P (\bar{A} \cap B) = P (B) ∴ P (A \cap B) = P (B) - P (A \cap B)$

For two events A and B,

Probability of occurrence of exactly are of A and B is $P (A \cap B) + P (A \cap \bar{B})$

From previous two questions,

P (A \cap B) = P (A) - P (A \cap B) and P (A \cap B) = P (B) - P (A \cap B) ∴ P (A \cap B) + P (A \cap B) = P (A) - P (A \cap B) + P (B) - P (A \cap B) = P (A) + P (B) - 2 P (A \cap B)

i.e Probability of occurrence of exactly are of A and B is $P (A) + P (B) - 2 P (A \cap B)$

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Question 13:

For two event A and B, if P(A) = P(A/B) = $\frac{1}{4} and P (B / A) = \frac{1}{2}$ , then A and B are _____________ events.

Answer:

For two events A and B
P(A) = P(A/B) = $\frac{1}{4} and P (B / A) = \frac{1}{2}$

$i . e \frac{P (A \cap B)}{P (B)} = \frac{1}{4} and \frac{P (A \cap B)}{P (A)} = \frac{1}{2} i . e P (A \cap B) = \frac{1}{4} P (B) i . e P (A \cap B) = P (A) P (B) [∵ P (A) = \frac{1}{4} given]$
∴ A and B are independent events.

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Question 14:

Let A and B be two events for which P(A) = a, P(B) = b, $P (A \cap B) = c, then P (A \cap B)$ = __________________.

Answer:

For two events A and B P(A) = a, P(B) = b, $P (A \cap B) = c$

Since $P (A \cap B) = P (B) - P (A \cap B)$

i.e

P (A \cap B) = b - c

Answer:

$Suppose S represents a student chosen randomly studying in class XII and G represents a female student chosen randomly . We have, P (G) = \frac{430}{1000} P (S / G) = \frac{43}{1000} Now, P (S / G) = \frac{P (S \cap G)}{P (G)} = \frac{\frac{43}{1000}}{\frac{430}{1000}} = \frac{1}{10}$

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Question 26:

Answer:

Let E₁, E₂ and E₃ denote the events of selecting Urn I, Urn II and Urn III, respectively.

Let A be the event that the two balls drawn are white and red.

$∴ P (E_{1}) = \frac{1}{3} P (E_{2}) = \frac{1}{3} P (E_{3}) = \frac{1}{3} Now, P (A / E_{1}) = \frac{^{1} C_{1} \times^{3} C_{1}}{^{6} C_{2}} = \frac{3}{15} = \frac{1}{5} P (A / E_{2}) = \frac{^{2} C_{1} \times^{1} C_{1}}{^{4} C_{2}} = \frac{2}{6} = \frac{1}{3} P (A / E_{3}) = \frac{^{4} C_{1} \times^{3} C_{1}}{^{12} C_{2}} = \frac{12}{66} = \frac{2}{11} Using Bayes' theorem, we get Required probability = P (E_{1} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) + + P (E_{3}) P (A / E_{3})} = \frac{\frac{1}{3} \times \frac{1}{5}}{\frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{11}} = \frac{\frac{1}{5}}{\frac{1}{5} + \frac{1}{3} + \frac{2}{11}} = \frac{\frac{1}{5}}{\frac{33 + 55 + 30}{165}} = \frac{33}{118} Required probability = P (E_{2} / A) = \frac{P (E_{2}) P (A / E_{2})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) + + P (E_{3}) P (A / E_{3})} = \frac{\frac{1}{3} \times \frac{1}{3}}{\frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{11}} = \frac{\frac{1}{3}}{\frac{1}{5} + \frac{1}{3} + \frac{2}{11}} = \frac{\frac{1}{3}}{\frac{33 + 55 + 30}{165}} = \frac{55}{118} Required probability = P (E_{3} / A) = \frac{P (E_{3}) P (A / E_{3})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) + + P (E_{3}) P (A / E_{3})} = \frac{\frac{1}{3} \times \frac{2}{11}}{\frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{11}} = \frac{\frac{2}{11}}{\frac{1}{5} + \frac{1}{3} + \frac{2}{11}} = \frac{\frac{2}{11}}{\frac{33 + 55 + 30}{165}} = \frac{30}{118}$

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Question 2:

A bag A contains 2 white and 3 red balls and a bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag B.

Let E₁ be the event that the outcome on the die is 1 or 2 and E₂ be the event that the outcome on the die is 3, 4, 5 or 6. Then,

$P (E_{1}) = \frac{2}{6} = \frac{1}{3} and P (E_{2}) = \frac{4}{6} = \frac{2}{3}$

Let A be the event of getting exactly one 'tail'.

P(A|E₁) = Probability of getting exactly one tail by tossing the coin three times if she gets 1 or 2 = $\frac{3}{8}$

P(A|E₂) = Probability of getting exactly one tail in a single throw of a coin if she gets 3, 4, 5 or 5 = $\frac{1}{2}$

As, the probability that the girl threw 3, 4, 5 or 6 with the die, if she obtained exactly one tail, is given by P(E₂|A).

So, by using Baye's theorem, we get

$P (E_{2} | A) = \frac{P (E_{2}) \times P (A | E_{2})}{P (E_{1}) \times P (A | E_{1}) + P (E_{2}) \times P (A | E_{2})} = \frac{(\frac{2}{3} \times \frac{1}{2})}{(\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2})} = \frac{(\frac{2}{6})}{(\frac{1}{8} + \frac{1}{3})} = \frac{(\frac{2}{6})}{(\frac{11}{24})} = \frac{24 \times 2}{11 \times 6} = \frac{8}{11}$

So, the probability that she threw 3, 4, 5 or 6 with the die if she obtained exactly one tail is $\frac{8}{11}$ .

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Question 6:

Two groups are competing for the positions of the Board of Directors of a Corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Answer:

Let E₁ and E₂ denote the events that the first group and the second group win the competition, respectively. Let A be the event of introducing a new product.

P(E₁) = Probability that the first group wins the competition = 0.6

P(E₂) = Probability that the second group wins the competition = 0.4

A factory has three machines X, Y and Z producing 1000, 2000 and 3000 bolts per day respectively. The machine X produces 1% defective bolts, Y produces 1.5% and Z produces 2% defective bolts. At the end of a day, a bolt is drawn at random and is found to be defective. What is the probability that this defective bolt has been produced by machine X?

Answer:

Let E₁, E₂ and E₃ denote the events that machine X produces bolts, machine Y produces bolts and machine Z produces bolts, respectively.

Let A be the event that the bolt is defective.

Total number of bolts = 1000 + 2000 + 3000 = 6000

P(E₁) = $\frac{1000}{6000} = \frac{1}{6}$

P(E₂) = $\frac{2000}{6000} = \frac{1}{3}$

P(E₃) = $\frac{3000}{6000} = \frac{1}{2}$

The probability that the defective bolt is produced by machine X is given by P (E₁/A).

$Now, P (A / E_{1}) = 1 % = \frac{1}{100} P (A / E_{2}) = 1.5 % = \frac{15}{1000} P (A / E_{3}) = 2 % = \frac{2}{100} Using Bayes' theorem, we get Required probability = P (E_{1} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) + + P (E_{2}) P (A / E_{2})} = \frac{\frac{1}{6} \times \frac{1}{100}}{\frac{1}{6} \times \frac{1}{100} + \frac{1}{3} \times \frac{15}{1000} + \frac{1}{2} \times \frac{2}{100}} = \frac{\frac{1}{6}}{\frac{1}{6} + \frac{1}{2} + 1} = \frac{\frac{1}{6}}{\frac{1 + 3 + 6}{6}} = \frac{1}{10}$

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Question 11:

An insurance company insured 3000 scooters, 4000 cars and 5000 trucks. The probabilities of the accident involving a scooter, a car and a truck are 0.02, 0.03 and 0.04 respectively. One of the insured vehicles meet with an accident. Find the probability that it is a (i) scooter (ii) car (iii) truck.

Answer:

Let E₁, E₂ and E₃ denote the events that the vehicle is a scooter, a car and a truck, respectively.

Let A be the event that the vehicle meets with an accident.

It is given that there are 3000 scooters, 4000 cars and 5000 trucks.

Total number of vehicles = 3000 + 4000 + 5000 = 12000

P(E₁) = $\frac{3000}{12000} = \frac{1}{4}$

P(E₂) = $\frac{4000}{12000} = \frac{1}{3}$

P(E₃) = $\frac{5000}{12000} = \frac{5}{12}$

The probability that the vehicle, which meets with an accident, is a scooter is given by P (E₁/A).

$Now, P (A / E_{1}) = 0.02 = \frac{2}{100} P (A / E_{2}) = 0.03 = \frac{3}{100} P (A / E_{3}) = 0.04 = \frac{4}{100} Using Bayes' theorem, we get (i) Required probability = P (E_{1} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) + + P (E_{2}) P (A / E_{2})} = \frac{\frac{1}{4} \times \frac{2}{100}}{\frac{1}{4} \times \frac{2}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{5}{12} \times \frac{4}{100}} = \frac{\frac{1}{2}}{\frac{1}{2} + 1 + \frac{5}{3}} = \frac{\frac{1}{2}}{\frac{3 + 6 + 10}{6}} = \frac{3}{19} (ii) Required probability = P (E_{2} / A) = \frac{P (E_{2}) P (A / E_{2})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) + + P (E_{2}) P (A / E_{2})} = \frac{\frac{1}{3} \times \frac{3}{100}}{\frac{1}{4} \times \frac{2}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{5}{12} \times \frac{4}{100}} = \frac{1}{\frac{1}{2} + 1 + \frac{5}{3}} = \frac{1}{\frac{3 + 6 + 10}{6}} = \frac{6}{19} (iii) Required probability = P (E_{2} / A) = \frac{P (E_{3}) P (A / E_{3})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) + + P (E_{2}) P (A / E_{2})} = \frac{\frac{5}{12} \times \frac{4}{100}}{\frac{1}{4} \times \frac{2}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{5}{12} \times \frac{4}{100}} = \frac{\frac{5}{3}}{\frac{1}{2} + 1 + \frac{5}{3}} = \frac{\frac{5}{3}}{\frac{3 + 6 + 10}{6}} = \frac{10}{19}$

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Question 12:

Suppose we have four boxes A, B, C, D containing coloured marbles as given below:
Figure

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A? box B? box C?

Answer:

Let R be the event of drawing the red marble.

Let E_A, E_B and E_C denote the events of selecting box A, box B and box C, respectively.

Total number of marbles = 40

Number of red marbles = 15

$∴ P (R) = \frac{15}{40} = \frac{3}{8}$

Probability of drawing a red marble from box A is given by P(E_A/R).

$∴ P (E_{A} / R) = \frac{P (E_{A} \cap R)}{P (R)} = \frac{\frac{1}{40}}{\frac{3}{8}} = \frac{1}{15}$

Probability of drawing a red marble from box B is given by P(E_B/R).

$∴ P (E_{B} / R) = \frac{P (E_{B} \cap R)}{P (R)} = \frac{\frac{6}{40}}{\frac{3}{8}} = \frac{2}{5}$

Probability of drawing a red marble from box C is given by P(E_C/R).

$∴ P (E_{C} / R) = \frac{P (E_{C} \cap R)}{P (R)} = \frac{\frac{8}{40}}{\frac{3}{8}} = \frac{8}{15}$

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Question 13:

A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job for 30% of the time and C on the job for 20% of the time. A defective item is produced. What is the probability that it was produced by A?

Let A be the event that the coin shows heads.

$∴ P (E_{1}) = \frac{1}{3} P (E_{2}) = \frac{1}{3} P (E_{3}) = \frac{1}{3} Now, P (A / E_{1}) = 1 P (A / E_{2}) = 75 % = \frac{3}{4} P (A / E_{2}) = \frac{1}{2} Using Bayes' theorem, we get Required probability = P (E_{1} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) + + P (E_{2}) P (A / E_{2})} = \frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}} = \frac{4}{9}$

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Question 28:

Assume that the chances of a patient having a heart attack is 40%. It is also assumed that meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options and patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Answer:

Let A, E₁ and E₂ denote the events that the selected person had a heart attack, did yoga and meditation, and followed the drug prescriptions, respectively.

$∴ P (E_{1}) = \frac{1}{2} P (E_{2}) = \frac{1}{2} Now, P (A / E_{1}) = 0.40 \times 0.70 = 0.28 P (A / E_{2}) = 0.40 \times 0.75 = 0.30 Using Bayes' theorem, we get Required probability = P (E_{1} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2})} = \frac{\frac{1}{2} \times 0.28}{\frac{1}{2} \times 0.28 + \frac{1}{2} \times 0.30} = \frac{14}{29}$

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Question 29:

Coloured balls are distributed in four boxes as shown in the following table:

Box	Colour
Box	Black	White	Red	Blue
I II III IV	3 2 1 4	4 2 2 3	5 2 3 1	6 2 1 5

A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III.

Answer:

Let A, E₁, E₂, E₃ and E₄ denote the events that the ball is black, box I selected, box II selected, box III is selected and box IV is selected, respectively.

$∴ P (E_{1}) = \frac{1}{4} P (E_{2}) = \frac{1}{4} P (E_{3}) = \frac{1}{4} P (E_{4}) = \frac{1}{4} Now, P (A / E_{1}) = \frac{3}{18} P (A / E_{2}) = \frac{2}{8} P (A / E_{3}) = \frac{1}{7} P (A / E_{4}) = \frac{4}{13} Using Bayes' theorem, we get Required probability = P (E_{3} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2})} = \frac{\frac{1}{4} \times \frac{1}{7}}{\frac{1}{4} \times \frac{3}{18} + \frac{1}{4} \times \frac{2}{8} + \frac{1}{4} \times \frac{1}{7} + \frac{1}{4} \times \frac{4}{13}} = \frac{\frac{1}{7}}{\frac{1}{6} + \frac{1}{4} + \frac{1}{7} + \frac{4}{13}} = \frac{156}{947}$

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Question 30:

If a machine is correctly set up it produces 90% acceptable items. If it is incorrectly set up it produces only 40% acceptable item. Past experience shows that 80% of the setups are correctly done. If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly set up.

Answer:

Let A be the event that the machine produces two acceptable items.
Also, let E₁ represent the event that the machine is correctly set up and E₂ represent the event that the machine is incorrectly set up

$∴ P (E_{1}) = 0.8 P (E_{2}) = 0.2 Now, P (A / E_{1}) = 0.9 \times 0.9 = 0.81 P (A / E_{2}) = 0.40 \times 0.40 = 0.16 Using Bayes' theorem, we get Required probability = P (E_{1} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2})} = \frac{0.8 \times 0.81}{0.8 \times 0.81 + 0.2 \times 0.16} = \frac{81}{85}$

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Question 31:

Bag A contains 3 red and 5 black balls, while bag B contains 4 red and 4 black balls. Two balls are transferred at random from bag A to bag B and then a ball is drawn from bag B at random. If the ball drawn from bag B is found to be red find the probability that two red balls were transferred from A to B.

Events E₁, E₂, E₃ and S be the events defined as follows:

E₁: The patient had disease d₁

E₂: The patient had disease d₂

E₃: The patient had disease d₃

S: The patient showed the symptom

E₁, E₂, and E₃ are mutually exclusive and exhaustive events.

$P (E_{1}) = \frac{1800}{5000} = \frac{18}{50} P (E_{2}) = \frac{2100}{5000} = \frac{21}{50} P (E_{3}) = \frac{1100}{5000} = \frac{11}{50}$
Now,

$P (\frac{S}{E_{1}})$ = Probability that the patient showed symptom given that patient had disease d₁= $\frac{1500}{5000} = \frac{15}{50}$

$P (\frac{S}{E_{2}})$ = Probability that the patient showed symptom given that patient had disease d₂= $\frac{1200}{5000} = \frac{12}{50}$

$P (\frac{S}{E_{3}})$ = Probability that the patient showed symptom given that patient had disease d₃= $\frac{900}{5000} = \frac{9}{50}$

Using Bayes theorem, we have

Probability that patient had disease d₁ such that symptom of d₁ showed = $P (\frac{E_{1}}{S}) = \frac{P (E_{1}) P (\frac{S}{E_{1}})}{P (E_{1}) P (\frac{S}{E_{1}}) + P (E_{2}) P (\frac{S}{E_{2}}) + P (E_{3}) P (\frac{S}{E_{3}})} = \frac{\frac{18}{50} \times \frac{15}{50}}{\frac{18}{50} \times \frac{15}{50} + \frac{21}{50} \times \frac{12}{50} + \frac{11}{50} \times \frac{9}{50}} = \frac{270}{621}$

Probability that patient had disease d₂ such that symptom of d₂ showed = $P (\frac{E_{2}}{S}) = \frac{P (E_{2}) P (\frac{S}{E_{2}})}{P (E_{1}) P (\frac{S}{E_{1}}) + P (E_{2}) P (\frac{S}{E_{2}}) + P (E_{3}) P (\frac{S}{E_{3}})} = \frac{\frac{21}{50} \times \frac{12}{50}}{\frac{18}{50} \times \frac{15}{50} + \frac{21}{50} \times \frac{12}{50} + \frac{11}{50} \times \frac{9}{50}} = \frac{252}{621}$

Probability that patient had disease d₃ such that symptom of d₃ showed = $P (\frac{E_{3}}{S}) = \frac{P (E_{3}) P (\frac{S}{E_{3}})}{P (E_{1}) P (\frac{S}{E_{1}}) + P (E_{2}) P (\frac{S}{E_{2}}) + P (E_{3}) P (\frac{S}{E_{3}})} = \frac{\frac{11}{50} \times \frac{9}{50}}{\frac{18}{50} \times \frac{15}{50} + \frac{21}{50} \times \frac{12}{50} + \frac{11}{50} \times \frac{9}{50}} = \frac{99}{621}$

Thus, the patient is most likely to have the disease d₁.

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Question 35:

A is known to speak truth 3 times out of 5 times. He throws a die and reports that it is one. Find the probability that it is actually one.

Answer:

Let A, E₁ and E₂ denote the events that the man reports the appearance of 1 on throwing a die, 1 occurs and 1 does not occur, respectively.

$∴ P (E_{1}) = \frac{1}{6} P (E_{2}) = \frac{5}{6} Now, P (A / E_{1}) = \frac{3}{5} P (A / E_{2}) = \frac{2}{5} Using Bayes' theorem, we get Required probability = P (E_{1} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2})} = \frac{\frac{1}{6} \times \frac{3}{5}}{\frac{1}{6} \times \frac{3}{5} + \frac{5}{6} \times \frac{2}{5}} = \frac{3}{3 + 10} = \frac{3}{13}$

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Question 36:

A speaks the truth 8 times out of 10 times. A die is tossed. He reports that it was 5. What is the probability that it was actually 5?

Answer:

Let A denote the event that man reports that 5 occurs and E the event that 5 actually turns up.

∴ P(E) = $\frac{1}{6}$ and $P (E) = 1 - \frac{1}{6} = \frac{5}{6}$

Also, $P (\frac{A}{E})$ = Probability that man reports that 5 occurs given that 5 actually turns up = Probability of man speaking the truth = $\frac{8}{10} = \frac{4}{5}$

$P (\frac{A}{E})$ = Probability that man reports that 5 occurs given that 5 doesnot turns up = Probability of man not speaking the truth = $1 - \frac{4}{5} = \frac{1}{5}$

∴ Required probability = $P (\frac{E}{A}) = \frac{P (E) P (\frac{A}{E})}{P (E) P (\frac{A}{E}) + P (E) P (\frac{A}{E})} = \frac{\frac{1}{6} \times \frac{4}{5}}{\frac{1}{6} \times \frac{4}{5} + \frac{5}{6} \times \frac{1}{5}} = \frac{4}{9}$

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Question 37:

In answering a question on a multiple choice test a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$ . What is the probability that a student knows the answer given that he answered it correctly?

Answer:

Let A, E₁ and E₂ denote the events that the answer is correct, the student knows the answer and the student guesses the answer, respectively.

$∴ P (E_{1}) = \frac{3}{4} P (E_{2}) = \frac{1}{4} Now, P (A / E_{1}) = 1 P (A / E_{2}) = \frac{1}{4} Using Bayes' theorem, we get Required probability = P (E_{1} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2})} = \frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}} = \frac{3}{3 + \frac{1}{4}} = \frac{12}{13}$

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Question 38:

A laboratory blood test is 99% effective in detecting a certain disease when its infection is present. 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 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Answer:

Let E₁ and E₂ denote the events that a person has a disease and a person has no disease, respectively.

E₁ and E₂ are complimentary to each other.

∴ P (E₁) + P (E₂) = 1

⇒ P (E₂) = 1 − P (E₁) = 1 − 0.001 = 0.999

Let A denote the event that the blood test result is positive.

$∴ P (E_{1}) = 0.1 % = 0.001 Now, P (A / E_{1}) = 99 % = 0.99 P (A / E_{2}) = 0.5 % = 0.005 Using Bayes' theorem, we get Required probability = P (E_{1} / A) = \frac{P (E_{1}) P (A / E_{1})}{P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2})} = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} = \frac{990}{5985} = \frac{22}{133}$

Page No 30.99:

Question 39:

There are three categories of students in a class of 60 students:
A : Very hardworking ; B : Regular but not so hardworking; C : Careless and irregular 10 students are in category A, 30 in category B and the rest in category C. It is found that the probability of students of category A, unable to get good marks in the final year examination is 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be one who could not get good marks in the examination. Find the probability that this student is category C.

Answer:

Let E denote the event that the student could not get good marks in the examination.

Also, A : the event that student is very hardworking

B : the event that student is regular but not so hardworking

C : the event that student is careless and irregular

∴ P(A) = $\frac{10}{60}$ , P(B) = $\frac{30}{60}$ and P(C) = $\frac{20}{60}$

Also,
$P (\frac{E}{A})$ = Pobability that the student of catagory A could not get good marks in the examination = 0.002

$P (\frac{E}{B})$ = Pobability that the student of catagory B could not get good marks in the examination = 0.02

$P (\frac{E}{C})$ = Pobability that the student of catagory C could not get good marks in the examination = 0.2

∴ Required probability = $P (\frac{C}{E}) = \frac{P (C) P (\frac{E}{C})}{P (A) P (\frac{E}{A}) + P (B) P (\frac{E}{B}) + P (C) P (\frac{E}{C})} = \frac{\frac{20}{60} \times 0.2}{\frac{10}{60} \times 0.002 + \frac{30}{60} \times 0.02 + \frac{20}{60} \times 0.2} = \frac{4}{4.62} = \frac{400}{462} = \frac{200}{231}$

Page No 31.70:

Question 33:

A card is drawn from a well-shuffled deck of 52 cards. The outcome is noted, the card is replaced and the deck reshuffled. Another card is then drawn from the deck.
(i) What is the probability that both the cards are of the same suit?
(ii) What is the probability that the first card is an ace and the second card is a red queen?

Answer:

$(i) P (both the cards are of same suit) = P (both the cards are of diamond) + P (both the cards are of spade) + P (both the cards are of club) + P (both the cards are of heart) = \frac{13}{52} \times \frac{13}{52} + \frac{13}{52} \times \frac{13}{52} + \frac{13}{52} \times \frac{13}{52} + \frac{13}{52} \times \frac{13}{52} = \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4} (ii) P (first ace and second red queen) = P (ace card) \times P (red queen) = \frac{4}{52} \times \frac{2}{52} = \frac{1}{338}$

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