### Probability Miscellaneous Exercise

## Question:

4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?

## Answer:

Step 1: Calculate the total number of possible outcomes. There are 52 cards in a deck, and 4 cards are drawn. Therefore, the total number of possible outcomes is 52 x 51 x 50 x 49 = 6,201,600.

Step 2: Calculate the number of favorable outcomes. There are 13 diamonds and 13 spades in a deck. Therefore, the number of favorable outcomes is 13 x 12 x 11 x 13 = 24,312.

Step 3: Calculate the probability. The probability of obtaining 3 diamonds and one spade is 24,312/6,201,600 = 0.0039 or 0.39%.

## Question:

In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets?

## Answer:

a) The probability of not getting a prize if you buy one ticket is 9,999/10,000 or 0.9999.

b) The probability of not getting a prize if you buy two tickets is 9,998/10,000 or 0.9998.

c) The probability of not getting a prize if you buy 10 tickets is 9,990/10,000 or 0.999.

## Question:

If 4-digit numbers greater than 5,000 are randomly formed from the digits 0,1,3,5 and 7. What is the probability of forming a number divisible by 5 when (i) the digits are repeated (ii) the repetition of digits is not allowed?

## Answer:

(i) When the digits are repeated:

The probability of forming a number divisible by 5 is 4/5.

This is because there are 4 possible combinations of the digits 0,1,3,5 and 7 that can be used to form a 4-digit number divisible by 5 (0001, 0035, 0351 and 3517). Since there are 5 possible combinations of the digits 0,1,3,5 and 7 that can be used to form a 4-digit number (0000, 0001, 0011, 0113, 0131, 0315, 0351, 0517, 1351, 1517, 3351, 3517, 5117 and 7111), the probability of forming a number divisible by 5 is 4/5.

(ii) When the repetition of digits is not allowed:

The probability of forming a number divisible by 5 is 2/5.

This is because there are 2 possible combinations of the digits 0,1,3,5 and 7 that can be used to form a 4-digit number divisible by 5 without repeating any of the digits (0035 and 0351). Since there are 5 possible combinations of the digits 0,1,3,5 and 7 that can be used to form a 4-digit number without repeating any of the digits (0113, 0131, 0315, 0517, 1351, 1517, 3351, 3517, 5117 and 7111), the probability of forming a number divisible by 5 is 2/5.

## Question:

A and B are two events such that P(A)=0.54,P(B)=0.69 and P(A ∩B) =0.35 Find (i) P(A∪B) (ii) P(A’ ∩B’) (iii) P(A ∩B’) (iv) P(B ∩A')

## Answer:

(i) P(A∪B) = P(A) + P(B) - P(A ∩B) = 0.54 + 0.69 - 0.35 = 0.88

(ii) P(A’ ∩B’) = 1 - (P(A) + P(B) - P(A ∩B)) = 1 - 0.88 = 0.12

(iii) P(A ∩B’) = P(A) - P(A ∩B) = 0.54 - 0.35 = 0.19

(iv) P(B ∩A’) = P(B) - P(A ∩B) = 0.69 - 0.35 = 0.34

## Question:

Three letters are dictated to three persons and an envelope is addressed to each of them the letters are inserted into the envelopes at random, so that each envelope contains exactly one letter. Find the probability that at least one letter is in proper envelope.

## Answer:

Answer: Step 1: The total number of possible outcomes is 3! = 6.

Step 2: The number of outcomes in which at least one letter is in the proper envelope is 3, since each letter can be put in the proper envelope.

Step 3: The probability that at least one letter is in the proper envelope is 3/6 = 1/2.

## Question:

A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once determine (i) P(2) (ii) P(1 or 3) (iii) P(not 3)

## Answer:

(i) P(2) = 3/6 = 1/2

(ii) P(1 or 3) = 2/6 + 1/6 = 3/6

(iii) P(not 3) = 5/6

## Question:

Out of 100 students two sections of 40 and 60 are formed. If you and your friend are among the 100 students. What is the probability that (a) you both enter the same sections? (b) you both enter the different sections?

## Answer:

(a) The probability that you and your friend both enter the same section is $\frac{40 \cdot 39}{100 \cdot 99}$

(b) The probability that you and your friend both enter different sections is $\frac{40 \cdot 60}{100 \cdot 99}$

## Question:

The number lock of a suitcase has 4 wheels each labelled with ten digits i.e. from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

## Answer:

Step 1: Calculate the total number of possible combinations. There are 10 digits (0-9) and 4 wheels, so the total number of possible combinations is 10 x 10 x 10 x 10 = 10,000.

Step 2: Calculate the number of combinations with no repeats. There are 10 choices for the first digit, 9 choices for the second digit (as it cannot be the same as the first), 8 choices for the third digit (as it cannot be the same as the first two), and 7 choices for the fourth digit (as it cannot be the same as the first three). So the total number of combinations with no repeats is 10 x 9 x 8 x 7 = 5040.

Step 3: Calculate the probability of getting the right sequence. The probability of getting the right sequence is the number of combinations with no repeats divided by the total number of possible combinations, which is 5040/10000 = 0.504 or 50.4%.

## Question:

A box contains 10 red marbles, 20 blue marbles and 30 green marbles, 5 marbles are drawn from the box what is the probability that (i) all will be blue? (ii) atleast one will be green?

## Answer:

(i) The probability that all 5 marbles will be blue is P(all blue) = (20/60) x (19/59) x (18/58) x (17/57) x (16/56) = 0.000717

(ii) The probability that at least one marble will be green is P(at least one green) = 1 - P(none green) = 1 - (40/60) x (39/59) x (38/58) x (37/57) x (36/56) = 0.9983

## Question:

From the employees of a company 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows: S.No Name Sex Age in years 1 Harish M 30 2 Rohan M 33 3 Sheetal F 46 4 Alis F 28 5 Salim M 41 A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years?

## Answer:

Step 1: Count the total number of people in the group: There are 5 people in the group.

Step 2: Count the number of males in the group: There are 3 males in the group.

Step 3: Count the number of people over 35 years in the group: There are 2 people over 35 years in the group.

Step 4: Calculate the probability: The probability that the spokesperson will be either male or over 35 years is 5/5 = 1.