### Probability Miscellaneous Exercise

## Question:

Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed ?

## Answer:

Step 1: Calculate the probability that exactly 6 people are right-handed. This can be done using the binomial probability formula:
P(x=6) = (10C6) * (0.9)^6 * (0.1)^{4}

Step 2: Calculate the probability that at most 6 people are right-handed. This can be done by adding the probabilities of 0, 1, 2, 3, 4, 5, and 6 people being right-handed: P(x<=6) = P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5) + P(x=6)

Step 3: Substitute the probabilities from Step 1 into Step 2 to calculate the probability that at most 6 people are right-handed:
P(x<=6) = (10C0) * (0.9)^{0} * (0.1)^{10} + (10C1) * (0.9)^{1} * (0.1)^{9} + (10C2) * (0.9)^{2} * (0.1)^{8} + (10C3) * (0.9)^{3} * (0.1)^{7} + (10C4) * (0.9)^{4} * (0.1)^{6} + (10C5) * (0.9)^{5} * (0.1)^{5} + (10C6) * (0.9)^{6} * (0.1)^{4}

Step 4: Simplify the expression to calculate the probability that at most 6 people are right-handed: P(x<=6) = 1 + 10 * 0.9 + 45 * 0.81 + 120 * 0.729 + 210 * 0.6561 + 252 * 0.59049 + 210 * 0.531441 = 0.9940

## Question:

An urn contains 25 balls of which 10 balls bear a mark ′X′ and the remaining 15 bear a mark ′Y′. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that (i) all will bear ′X′ mark (ii) not more than 2 will bear ′Y′ mark (iii) at least one ball will bear ′Y′ mark (iv) the number of balls with ′X′ mark and ′Y′ mark will be equal.

## Answer:

(i) The probability that all 6 balls drawn will bear the ‘X’ mark is (10/25)^{6}.

(ii) The probability that not more than 2 of the 6 balls drawn will bear the ‘Y’ mark is (15/25)^{2} * (10/25)^4.

(iii) The probability that at least one ball will bear the ‘Y’ mark is 1 - (10/25)^{6}.

(iv) The probability that the number of balls with ‘X’ mark and ‘Y’ mark will be equal is (15/25)^{3} * (10/25)^{3}.

## Question:

An electronic assembly consists of two subsystems, say A and B. From previous testing procedures, the following probabilities are assumed to be known: P(A) fails =0.2 P(B) fails alone =0.15P(A and B) failing =0.15 Evaluate the following probabilities: (i) P(A fails or B fails) (ii) P (A fails alone)

## Answer:

(i) P(A fails or B fails) = P(A fails) + P(B fails) - P(A and B fails) = 0.2 + 0.15 - 0.15 = 0.2

(ii) P(A fails alone) = P(A fails) - P(A and B fails) = 0.2 - 0.15 = 0.05

## Question:

A couple has two children, (i) Find the probability that both children are males, if it is known that at least one of the children is male. (ii) Find the probability that both children are females, if ti is known that the elder child is a female. A 0.55,0.38 B 0.33,0.50 C 0.67,0.78, D 0.56,0.67

## Answer:

Answer: D 0.56,0.67

## Question:

Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce 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 the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

## Answer:

Given: P(Heart Attack) = 40% P(Meditation and Yoga) = 30% P(Drug Prescription) = 25%

P(Patient followed Meditation and Yoga) = 50%

To find: P(Patient followed Meditation and Yoga | Heart Attack)

Using Bayes’ Theorem:

P(Patient followed Meditation and Yoga | Heart Attack) = P(Heart Attack | Patient followed Meditation and Yoga) * P(Patient followed Meditation and Yoga) / P(Heart Attack)

P(Heart Attack | Patient followed Meditation and Yoga) = (1 - P(Meditation and Yoga)) = (1 - 30%) = 70%

P(Patient followed Meditation and Yoga | Heart Attack) = 70% * 50% / 40%

P(Patient followed Meditation and Yoga | Heart Attack) = 58.33%

## Question:

An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.

## Answer:

Step 1: Calculate the probability of success in a single trial. P(success) = 2/3

Step 2: Calculate the probability of 4 successes in 6 trials.
P(4 successes in 6 trials) = (2/3)^{4} x (1/3)^{2}

Step 3: Calculate the probability of at least 4 successes in 6 trials. P(at least 4 successes in 6 trials) = 1 - P(3 successes in 6 trials)

Step 4: Calculate the probability of 3 successes in 6 trials.
P(3 successes in 6 trials) = (2/3)^{3} x (1/3)^{3}

Step 5: Calculate the probability of at least 4 successes in 6 trials.
P(at least 4 successes in 6 trials) = 1 - (2/3)^{3} x (1/3)^{3}

Answer: P(at least 4 successes in 6 trials) = 0.868

## Question:

Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male ? Assume that there are equal number of males and females.

## Answer:

Answer: Step 1: Calculate the total percentage of people with grey hair. 5% of men + 0.25% of women = 5.25%

Step 2: Calculate the probability of the selected person being male. 5% / 5.25% = 0.9615384615384616

## Question:

A die is thrown again and again until three sixes are obtained. Find a if the probability of obtaining the third six in the sixth throw of the die is a/23328.

## Answer:

Step 1: The probability of obtaining the third six in the sixth throw of the die is 1/216.

Step 2: a = 1/216 x 23328 = 108

## Question:

If P(A∣B)>P(A), then which of the following is correct A P(B∣A)<P(B) B P(A∩B)<P(A)⋅P(B) C P(B∣A)>P(B) D P(B∣A)=P(B)

## Answer:

Answer: C

Explanation: Given: P(A|B) > P(A)

We can use the formula P(A|B) = P(A∩B)/P(B)

Since P(A|B) > P(A), then P(A∩B) > P(A)⋅P(B).

Therefore, Option C is correct: P(B|A) > P(B).

## Question:

What is the minimum number of times must a man toss a fair coin so that the probability of having at least one head is more than 90% ?

## Answer:

Answer:

Step 1: Understand the question: The question is asking for the minimum number of times a man must toss a fair coin so that the probability of having at least one head is more than 90%.

Step 2: Calculate the probability of having at least one head in each toss: The probability of having at least one head in a single toss is 50% (since it is a fair coin).

Step 3: Calculate the minimum number of tosses required to have a probability of more than 90%: To have a probability of more than 90%, we need to calculate the cumulative probability of having at least one head in multiple tosses. This can be done using the formula P(X≥1) = 1 - P(X=0).

Therefore, the minimum number of tosses required to have a probability of more than 90% is 10 tosses.

## Question:

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I and Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find 310 times the probability that the transferred ball is black.

## Answer:

Step 1: Find the probability that the ball drawn from Bag II is red.

Number of red balls in Bag II = 4 Total number of balls in Bag II = 9

Probability that the ball drawn from Bag II is red = 4/9

Step 2: Find the probability that the transferred ball is black.

Number of black balls in Bag I = 4 Total number of balls in Bag I = 7

Probability that the transferred ball is black = 4/7

Step 3: Calculate 310 times the probability that the transferred ball is black.

310 times the probability that the transferred ball is black = 310 x (4/7) = 440/7

## Question:

In a hurdle race, a player has to cross 10 hurdles. The probability that he will clear each hurdle is 5/6. What is the probability that he will knock down fewer than 2 hurdles ?

## Answer:

Answer: Step 1: Calculate the probability of clearing 8 hurdles (P(8)) P(8) = (5/6)^8

Step 2: Calculate the probability of clearing 9 hurdles (P(9)) P(9) = (5/6)^9

Step 3: Calculate the probability of clearing 10 hurdles (P(10)) P(10) = (5/6)^10

Step 4: Calculate the probability of knocking down fewer than 2 hurdles P(knocking down fewer than 2 hurdles) = 1 - (P(8) + P(9) + P(10)) = 1 - ((5/6)^8 + (5/6)^9 + (5/6)^10) = 0.871

## Question:

In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he win / loses.

## Answer:

Answer: Step 1: Calculate the probability of getting a six in each throw. P(getting a six) = 1/6

Step 2: Calculate the expected value of the amount won/lost in each throw. Expected value of each throw = (1 rupee x probability of getting a six) - (1 rupee x probability of not getting a six) = (1 rupee x 1/6) - (1 rupee x 5/6) = 1/6 - 5/6 = -4/6

Step 3: Calculate the expected value of the amount won/lost in all three throws. Expected value of all three throws = (-4/6) x 3 = -12/6 = -2 rupees

## Question:

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 1/2).

## Answer:

Answer:

- The determinant in question has two rows and two columns, with each element being either 0 or 1.
- The value of the determinant can be either 0, 1, or -1.
- The probability of the determinant being positive is equal to the probability that the value of the determinant is 1.
- The probability of the value of the determinant being 1 is equal to the probability of both elements in the first row being 1 multiplied by the probability of both elements in the second row being 1.
- The probability of both elements in the first row being 1 is (1/2) × (1/2) = 1/4.
- The probability of both elements in the second row being 1 is (1/2) × (1/2) = 1/4.
- Therefore, the probability of the value of the determinant being positive is equal to the product of the probabilities of both rows being 1, which is 1/4 × 1/4 = 1/16.

## Question:

If A and B are two events such that P(A)=0 and P(B∣A)=1, then A A⊂B B B⊂A C B=ϕ D A=ϕ

## Answer:

Answer: B B⊂A

## Question:

If A and B are any two events such that P(A)+P(B)−P(AandB)=P(A), then A P(B∣A)=1 B P(A∣B)=1 C P(B∣A)=0 D P(A∣B)=0

## Answer:

A) False B) False C) True D) True