### Probability Exercise 05

## Question:

The probability that a student is not a swimmer is 51. Then the probability that out of five students, four are swimmer is
A 5C4(4/5)^{4} 1/5
B (4/5)^{4} 1/5
C 5C1 1/5(4/5)^{4}
D None of these

## Answer:

Answer: C 5C1 1/5(4/5)^{4}

## Question:

A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

## Answer:

Answer: Step 1: Calculate the total number of possible outcomes when four balls are drawn successively with replacement from the bag.

Since there are 10 balls in the bag and each ball can be chosen more than once, the total number of possible outcomes is 10x10x10x10 = 10,000.

Step 2: Calculate the number of possible outcomes where none of the four balls is marked with the digit 0.

Since there are 9 balls that are not marked with the digit 0, the number of possible outcomes where none of the four balls is marked with the digit 0 is 9x9x9x9 = 6,561.

Step 3: Calculate the probability that none of the four balls is marked with the digit 0.

The probability that none of the four balls is marked with the digit 0 is 6,561/10,000 = 0.6561 or 65.61%.

## Question:

Suppose X has a binomial distribution B of 6,1/2. Show that X=3 is the most likely outcome. (Hint : P(X=3) is the maximum among all P(xi),xi=0,1,2,3,4,5,6)

## Answer:

Step 1: A binomial distribution B of 6,1/2 implies that the random variable X has 6 trials and the probability of success is 1/2.

Step 2: Calculate the probability of each outcome, P(X=0), P(X=1), P(X=2), P(X=3), P(X=4), P(X=5), and P(X=6).

Step 3: Compare the probabilities of each outcome and determine which is the most likely.

Step 4: The most likely outcome is X=3 since P(X=3) is the maximum among all P(xi),xi=0,1,2,3,4,5,6.

## Question:

The probability of throwing at most 2 sixes in 6 throws of a single die is =a/b⋅(5/6)^{4}. Find a+b

## Answer:

a+b = 6

## Question:

In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers ’true’, if it falls tails, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.

## Answer:

Answer: Step 1: Find the probability of answering one question correctly. P(answering one question correctly) = P(heads) = 0.5

Step 2: Find the probability of answering 12 questions correctly. P(answering 12 questions correctly) = P(heads)^12 = 0.5^12 = 0.000244140625

Step 3: Find the probability of answering at least 12 questions correctly. P(answering at least 12 questions correctly) = 1 - P(answering less than 12 questions correctly) = 1 - (1 - P(answering 12 questions correctly)) = 1 - (1 - 0.000244140625) = 0.999755859375

## Question:

The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs. (i) none (ii) not more than one (iii) more than one (iv) at least one will fuse after 150 days of use.

## Answer:

(i) none: 0.95^5 = 0.7737

(ii) not more than one: 0.05 * 5 + 0.95^{4} = 0.9223

(iii) more than one: 1 - 0.9223 = 0.0777

(iv) at least one: 1 - 0.7737 = 0.2263

## Question:

Find the probability of getting 5 exactly twice in 7 throws of a die.

## Answer:

Step 1: Calculate the total number of possible outcomes in 7 throws of a die. This can be done by multiplying the number of possible outcomes in each throw, which is 6. Therefore, the total number of possible outcomes is 6 x 6 x 6 x 6 x 6 x 6 x 6 = 279,936.

Step 2: Calculate the number of ways to get 5 exactly twice in 7 throws of a die. This can be done by multiplying the number of ways to get 5 in each throw, which is 1, and the number of ways to get any other number in each throw, which is 5. Therefore, the number of ways to get 5 exactly twice in 7 throws of a die is 5 x 5 x 5 x 5 x 5 x 1 x 1 = 15625.

Step 3: Calculate the probability of getting 5 exactly twice in 7 throws of a die. This can be done by dividing the number of ways to get 5 exactly twice in 7 throws of a die (15625) by the total number of possible outcomes in 7 throws of a die (279,936). Therefore, the probability of getting 5 exactly twice in 7 throws of a die is 0.0557.

## Question:

A pair of dice is thrown 4 times. If getting a double is considered a success, find the probability of two successes(approximately). A 0.12 B 0 C 0.86 D 1

## Answer:

Answer: C 0.86

## Question:

A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of (i) 5 successes ? (ii) at least 5 successes ? (iii) at most 5 successes ?

## Answer:

(i) The probability of 5 successes is (1/2)^{5} * (1/2) = 1/32.

(ii) The probability of at least 5 successes is 1 - (1/2)^{5} = 31/32.

(iii) The probability of at most 5 successes is (1/2)^{5} + (1/2)^{4} + (1/2)^{3} + (1/2)^{2} + (1/2) + 1 = 31/32.

## Question:

It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective ?

## Answer:

Step 1: Calculate 10% of 12 articles, which is 1.2.

Step 2: Calculate the probability of 9 out of 12 articles being defective. This can be done by finding the binomial coefficient for 9 and 12, which is 924.

Step 3: Divide the binomial coefficient by the total number of possible combinations, which is 479001600.

Step 4: Multiply the result by 1.2 to get the probability of 9 out of 12 articles being defective.

Answer: The probability that in a random sample of 12 such articles, 9 are defective is 0.0019379845.

## Question:

There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item ?

## Answer:

- Calculate the total number of items in the large bulk:

Let n = the total number of items in the large bulk

5% of n = 0.05n

0.05n = defective items

n = 0.05n / 0.05

n = 1 / 0.05

n = 20

- Calculate the number of defective items in the sample of 10 items:

Let x = the number of defective items in the sample of 10 items

5% of 10 = 0.05 x 10

0.05 x 10 = defective items

x = 0.05 x 10 / 0.05

x = 10 / 0.05

x = 200

- Calculate the probability that a sample of 10 items will include not more than one defective item:

The probability that a sample of 10 items will include not more than one defective item =

P(x ≤ 1) = P(x = 0) + P(x = 1)

P(x = 0) = (20C0) (0.95)^{10} (0.05)^0

P(x = 0) = 1 (0.95)^{10} (1)

P(x = 0) = 0.5987

P(x = 1) = (20C1) (0.95)^{9} (0.05)^{1}

P(x = 1) = 20 (0.95)^{9} (0.05)

P(x = 1) = 0.3829

P(x ≤ 1) = 0.5987 + 0.3829

P(x ≤ 1) = 0.9816

## Question:

Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that (i) all the five cards are spades ? (ii) only 3 cards are spades ? (iii) none is a spade ?

## Answer:

(i) The probability that all the five cards are spades is (1/4)^{5} = 1/1024.

(ii) The probability that only 3 cards are spades is (1/4)^{3} x (3/4)^{2} = 27/1024.

(iii) The probability that none of the cards is a spade is (3/4)^{5} = 243/1024.

## Question:

On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing ?

## Answer:

Answer: Step 1: Calculate the total number of possible outcomes. There are 3 possible answers for each of the 5 questions, so the total number of possible outcomes is 3 x 3 x 3 x 3 x 3 = 243.

Step 2: Calculate the number of outcomes where the candidate gets 4 or more correct answers. There are 3 possible outcomes for each question, so if the candidate correctly guesses 4 or more of the questions, there are 3 x 3 x 3 x 3 = 81 possible outcomes.

Step 3: Calculate the probability. The probability of getting 4 or more correct answers just by guessing is 81/243 = 0.33 or 33%.

## Question:

A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1/100. What is the probability that he will win a prize (a) at least once (b) exactly once (c) at least twice ?

## Answer:

a) The probability that he will win a prize at least once is 1 - (99/100)^{50} = 0.632

b) The probability that he will win a prize exactly once is (1/100) * (99/100)^{49} = 0.315

c) The probability that he will win a prize at least twice is 1 - (99/100)^{49} - (1/100) * (99/100)^{49} = 0.317