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- Most Important Multiple Choice Questions
- Online Probability Exercise with Correct Answer Key and Solutions
- Useful for all Competitive Exams

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- Question 1 of 20
##### 1. Question

Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is

##### Hint

- Question 2 of 20
##### 2. Question

The probability that A speaks truth is ⅘ while this probability for B is ¾. The probability that they contradict each other when asked to speak on a fact is

##### Hint

A and B will contradict each other if one speaks truth and other false . So, the required

Probability

- Question 3 of 20
##### 3. Question

The probability that a student is not a swimmer is ⅕. Then the probability that out of the five students, four are swimmers is :

##### Hint

4 students out of 5 can be selected in ⁵C₄ ways.

Probability of a student being not a swimmer ⅕

Probability of a student being a swimmer

Required probability

- Question 4 of 20
##### 4. Question

I forgot the last digit of a 7-digit telephone number. If I randomly dial the final 3 digits after correctly dialling the first four, then what is the chance of dialling the correct number?

##### Hint

It is given that last 3 digits are randomly dialled. Then each of the digits can be selected out of 10 digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) in 10 ways.

Hence the required probability

- Question 5 of 20
##### 5. Question

If two squares are chosen at random on a chess board, the probability that they have a side in common is

##### Hint

Two squares out of 64 can be selected in

The number of ways of selecting those pairs which have a side in common

= ½ (4×2+24×3+36×4) = 112

[Since each of the corner squares has two neighbours each of 24 squares in border rows, other than corner ones has three neighbours and each of the remaining 36 squares have four neighbours and in this computation, each pair of squares has been considered twice].

Hence required probability

=

- Question 6 of 20
##### 6. Question

A die is loaded such that the probability of throwing the number i is proportional to its reciprocal. The probability that 3 appears in a single throw is :

##### Hint

- Question 7 of 20
##### 7. Question

There is a five-volume dictionary among 50 books arranged on a shelf in random order. If the volumes are not necessarily kept side by side, the probability that they occur in increasing order from left to right is :

##### Hint

The number of ways of arranging 50 books

= ⁵⁰P₅₀ = 50!.

The number of ways of choosing places for the five volume dictionary

= ⁵⁰C₅

and the number of ways of arranging the remaining 45 books

= ⁴⁵C₄₅= (45)!

Thus the number of favourable ways

= (⁵⁰C₅) (45 !).

Hence the probability of the required event

- Question 8 of 20
##### 8. Question

Atul can hit a target 3 times in 6 shots, Bhola can hit the target 2 times in 6 shots and Chandra can hit the 4 times in 4 shots. What is the probability that at least 2 shots (out of 1 shot taken by each one of them) hit the target?

##### Hint

Chandra hits the target 4 times in 4 shots. Hence, he hits the target definitely.

The required probability, therefore, is given by.

P(both Atul and Bhola hit) + P(Atul hits, Bhola does not hit) + P(Atul does not hit, Bhola hits)

=

- Question 9 of 20
##### 9. Question

A bag contain 5 white, 7 red and 8 black balls. If 4 balls are drawn one by one with replacement, what is the probability that all are white?

##### Hint

Total number of balls

= 5 + 7 + 8 = 20

Probability that the first ball drawn is white

If balls are drawn with replacement, all the four events will have equal probability.

Therefore, required probability

- Question 10 of 20
##### 10. Question

A dice is thrown 6 times. If ‘getting an odd number’ is a ‘success’, the probability of 5 successes is :

##### Hint

Let A be the event of getting an odd number.

Here, n (S) = 6 and

n (A) = 3

Probability of getting an odd number

Hence, probability of not getting an odd number

Required probability of 5 successes

- Question 11 of 20
##### 11. Question

There are 6 positive and 8 negative numbers. Four numbers are chosen at random and multiplied. The probability that the product is a positive number is:

##### Hint

Total no. of numbers

= 6 positive + 8 negative = 14

n(S) = ¹⁴C₄

The product of four numbers could be positive when,

(a) all the four numbers chosen are positive or

(b) all the four numbers chosen are negative or

(c) two of the chosen numbers are positive and two are negative.

Required Prob.

=

=

- Question 12 of 20
##### 12. Question

Two dice are tossed. The probability that the total score is a prime number is :

##### Hint

Total no. of outcomes when two dice are thrown = n (S) = 36 and the possible cases for the event that the sum of numbers on two dice is a prime number, are

(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 1), (5, 6), (6, 1), (6, 5)

Number of outcomes favouring the event

= n (A) = 15

Required probability

- Question 13 of 20
##### 13. Question

In his wardrobe, Timothy has 3 trousers. One of them is black and second blue, and the third brown. In the wardrobe, he also has 4 shirts. One of them is black and the other 3 are white. He opens his wardrobe in the dark and picks out one shirt-trouser pair without examining their colour. What is the likelihood that neither the shirt nor the trouser is black?

##### Hint

Probability that the trouser is not black = ⅔

Probability that the shirt is not black = ¾

Since, picking of a shirt and a trouser are independent,

required probability

- Question 14 of 20
##### 14. Question

A car is parked by an owner amongst 25 cars in a row, not at either end. On his return he finds that exactly 15 places are still occupied. The probability that both the neighbouring places are empty is

##### Hint

Exhaustive number of cases = ²⁴C₁₄

Favourable cases = ²²C₁₄

- Question 15 of 20
##### 15. Question

The letters of the word SOCIETY are placed at random in a row. The probability that the three vowels come together is

##### Hint

The word ‘society’ contains seven distinct letters and they can be arranged at random in a row in

⁷P₇ ways, i.e. in 7! = 5040 ways.

Let us now consider those arrangements in which all the three vowels come together. So in this case we have to arrange four letters. S,C,T,Y and a pack of three vowels in a row which can be done in

⁵ P₅ i.e. 5! = 120 ways.

Also, the three vowels in their pack can be arranged in

³P₃ i.e. 3! = 6 ways.

Hence, the number of arrangements in which the three vowels come together is

120 × 6 = 720

∴ The probability that the vowels come together

=

- Question 16 of 20
##### 16. Question

The probability that the 13th day of a randomly chosen month is a Friday, is

##### Hint

Probability of selecting a month = ¹⁄₁₂.

13th day of the month is Friday if its first day is Sunday and the probability of this = ¹⁄₇.

∴ Required probability

=.

- Question 17 of 20
##### 17. Question

If a leap year selected at random, the chance that it will contain 53 Sunday is

##### Hint

A leap-year has 366 days i.e. 52 complete weeks and two days more these two days be two consecutive days of a week. A leap year will have 53 Sundays if out of the two consecutive days of a week selected at random one is a Sunday.

Let S be the sample space and E be the event that out of the two consecutive days of a week one is Sunday, then

S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)}

∴ n (S) = 7 and E = {(Sunday, Monday), (Saturday, Sunday)

∴ n (E) = 2

Now, required Probability, P(E) = = ²⁄₇

- Question 18 of 20
##### 18. Question

A and B toss a fair coin each simultaneously 50 times. The probability that both of them will not get tail at the same toss is

##### Hint

For each toss there are four choices

(i) A gets head, B gets head

(ii) A gets tail, B gets head

(iii) A gets head, B gets tail

(iv) A gets tail, B gets tail

thus, exhaustive number of ways

= 4⁵⁰.

Out of the four choices listed above (iv) is not favourable to the required event in a toss.

Therefore favourable number of cases is 3⁵⁰.

Hence, the required probability =

- Question 19 of 20
##### 19. Question

Suppose six coins are tossed simultaneously. Then the probability of getting at least one tail is :

##### Hint

If six coins are tossed, then the total no. of outcomes

= (2)⁶ = 64

Now, probability of getting no tail = ¹⁄₆₄

Probability of getting at least one tail

- Question 20 of 20
##### 20. Question

A coin is tossed 5 times. What is the probability that head appears an odd number of times?

##### Hint

Probability of occurrence of head in a toss of a coin is 1/2.

Required probability = Prob. [Head appears once] + Prob.[Head appears thrice] + Prob. [Head appears five times]

=

= [5 + 10 + 1]

=

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