Probability of Random Experiments MCQ Quiz - Objective Question with Answer for Probability of Random Experiments - Download Free PDF

Last updated on Jul 8, 2025

Latest Probability of Random Experiments MCQ Objective Questions

Probability of Random Experiments Question 1:

An experiment yields 3 mutually exclusive and exhaustive events A, B, and C.

If P(A) = 2P(B) = 3P(C), then P(A) is

  1. 2/11
  2. 3/11
  3. 5/11
  4. 6/11

Answer (Detailed Solution Below)

Option 4 : 6/11

Probability of Random Experiments Question 1 Detailed Solution

Given, P(A) = 2P(B) = 3P(C)

⇒ P(C) = 2/3 P(B)

Since A, B, C are three mutually exclusive and exhaustive events

∴ P(A) + P(B) + P(C) = 1

⇒ P(B) = 3/11

From the given relation P(A) = 2 P(B) = 6/11

Probability of Random Experiments Question 2:

If two fair dice are tossed, then what is the probability that the sum of the numbers on the faces of the dice is strictly greater than 7?

  1. 1/3
  2. 5/12
  3. 7/12
  4. 3/4

Answer (Detailed Solution Below)

Option 2 : 5/12

Probability of Random Experiments Question 2 Detailed Solution

Calculation:

Given,

Two fair six-sided dice are rolled.

Each die has faces numbered 1 through 6.

Total number of outcomes:

Since each die has 6 faces, the sample space size is

.

Favourable outcomes (sum > 7):

We list all ordered pairs ((i, j)) with (i + j > 7):

Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes

Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes

Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes

Sum = 11: (5,6), (6,5) → 2 outcomes

Sum = 12: (6,6) → 1 outcome

Total favourable outcomes = (5 + 4 + 3 + 2 + 1 = 15).

Probability calculation:

The probability that the sum is strictly greater than 7 is

7) = \frac{\text{favourable outcomes}}{\text{total outcomes}} = \frac{15}{36} = \frac{5}{12}\).

Hence, the correct answer is Option 2.

Probability of Random Experiments Question 3:

Let N be the sum of the numbers appeared when two fair dice are rolled and let the probability that   are in geometric progression be . Then the value of k is

  1. 2
  2. 4
  3. 16
  4. 8

Answer (Detailed Solution Below)

Option 2 : 4

Probability of Random Experiments Question 3 Detailed Solution

Calculation: 

Given :  are in G.P. 

n(s) = 36

⇒ 3N = (N – 2)(N + 2)

⇒ 3N = N2 – 4

⇒ N2 – 3N – 4 = 0

⇒ (N – 4)(N + 1) = 0 ⇒ N 4 or N = –1 rejected

⇒ (Sum = 4) ≡ {(1, 3), (3,1), (2,2)}

⇒ n(A) = 3 

⇒ 

Hence, the correct answer is Option 2.

Probability of Random Experiments Question 4:

A bag is Randomly selected, If drawn ball is red, then probability that ball is selected from bag-I is p. If ball drawn is green then probability that ball is selected from bag-III is q. Then  equals to

  Red  Blue  Green 
Bag-I  3 3 4
Bag-II  4 3 3
Bag-III  5 2 3

Answer (Detailed Solution Below)

Option 1 :

Probability of Random Experiments Question 4 Detailed Solution

Answer (1)

Sol.

p(B/ R) = 

p(BG) = 

Probability of Random Experiments Question 5:

Let U1, U2, ..., U5 be 5 urns such that urn U contains 2k + k2 balls, out of which 2k are white balls and k2 are black balls, k = 1, 2,..., 5. An urn is selected with probability of selecting urn Uk being proportional to (k+2). A ball is chosen randomly from the selected urn. Then, the probability that the urn U5 was selected, given that the ball drawn is white, is equal to

Answer (Detailed Solution Below)

Option 3 :

Probability of Random Experiments Question 5 Detailed Solution

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Top Probability of Random Experiments MCQ Objective Questions

A bag contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. The probability of getting the balls of different colors is:

  1. None of these

Answer (Detailed Solution Below)

Option 2 :

Probability of Random Experiments Question 6 Detailed Solution

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Concept:

  • The probability of drawing ‘k objects of type p’ from a collection of n = p + q + r + … objects is, given as: .
  • Probability of a Compound Event [(A and B) or (B and C)] is calculated as:

    P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]

    ('and' means '×' and 'or' means '+')

 

Calculation:

There are a total of 7 red + 4 blue = 11 balls.

Probability of drawing 1 red ball = .

Probability of drawing 1 blue ball = .

Probability of drawing (1 red) AND (1 blue) ball = .

Similarly, Probability of drawing (1 blue) AND (1 red) ball = .

Probability of getting the balls of different colors =  +  = 

A and B are two events such that P(B) = 0.4 and P(A ∪ B) = 0.6 If A and B are independent, then P(A) is

Answer (Detailed Solution Below)

Option 2 :

Probability of Random Experiments Question 7 Detailed Solution

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Concept:

Independent events:

Two events are independent if the incidence of one event does not affect the probability of the other event.

If A and B are two independent events, then P(A ∩ B) = P(A) × P(B)

 

Calculation:

Given: P(B) = 0.4 and P(A ∪ B) = 0.6

P(A ∪ B) = 0.6

⇒ P(A) + P(B) - P(A ∩ B) = 0.6

⇒ P(A) + P(B) - P(A) × P(B) = 0.6               (∵ A and B are independent events.)

⇒ P(B) + P(A) [1 - P(B)] = 0.6

⇒ 0.4 + P(A) [1 - 0.4] = 0.6

⇒ P(A) × 0.6 = 0.2 

In a room there are eight couples. Out of them if 4 people are selected at random, the probability that they may be couples is

Answer (Detailed Solution Below)

Option 4 :

Probability of Random Experiments Question 8 Detailed Solution

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Concept:

1) Combination: Selecting r objects from given n objects.

  • The number of selections of r objects from the given n objects is denoted by 


2) Probability of an event happening = 

Note: Use combinations if a problem calls for the number of ways of selecting objects.

Calculation:

Given:

In a room, there are eight couples.

⇒ Eight couples = 16 peoples

We have to select four peoples out of 16 peoples.

⇒ Total possible cases = 16C4

Now, we have to select four people- they may be couples

So, we have to select two couples from eight couples.

⇒ Favourable cases = 8C2

Hence Required Probability = 

Three mangoes and three apples are in box. If two fruits are chosen at random, the probability that one is a mango and the other is an apple is

  1. 2/3
  2. 3/5
  3. 1/3
  4. 4/5

Answer (Detailed Solution Below)

Option 2 : 3/5

Probability of Random Experiments Question 9 Detailed Solution

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Concept:

If S is a sample space and E is a favourable event then the probability of E is given by:

Calculation:

Total fruits = 3 + 3 = 6

Total possible ways = 6C2 = 15 = n(S)

Favourable ways = 3C1 × 3C1 = 9 = n(E)

∴ Required probability =

An unbiased coin is tossed 3 times, if the third toss gets head what is the probability of getting at least one more head?

Answer (Detailed Solution Below)

Option 1 :

Probability of Random Experiments Question 10 Detailed Solution

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Concept: 

  • The number of ways for selecting r from a group of n (n > r) = nCr 
  • The probability of particular case

Calculation

If it is known that third toss gets head, the possible cases:

(H, H, H), (H, T, H), (T, H, H), (T, T, H)

∴ Total cases possible = 4

Total favourable cases = 3 [(H, H, H), (H, T, H), (T, H, H)]

So, required probability P = 

P = 

If a coin is tossed thrice, find the probability of getting one or two heads.

Answer (Detailed Solution Below)

Option 3 :

Probability of Random Experiments Question 11 Detailed Solution

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Concept:

P(A) = 

Where n(A) = No. of favourable cases for event A and n(S) = cardinality of sample space.

Solution:

If a coin is tossed thrice, possible outcomes are:

S = {HHH, HHT, HTH, THH, THT, TTH, HTT, TTT}

Probability of getting one or two heads:

A = {HHT, HTH, THH, THT, TTH, HTT}

The number of possible outcomes, when a coin is tossed 6 times, is

  1. 36
  2. 64
  3. 12
  4. None of these

Answer (Detailed Solution Below)

Option 2 : 64

Probability of Random Experiments Question 12 Detailed Solution

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Concept: 

Sample space is nothing but a set of all possible outcomes of the experiment.

If we toss a coin n times then possible outcomes or number of elements in sample space = 2n elements

Calculation:

Number of outcomes when a coin is tossed = 2 (Head or Tail)

∴Total possible outcomes when a coin is tossed 6 times = 2 ×  2 × 2 × 2 × 2 × 2 = 64

If four dice are thrown together, then what is the probability that the sum of the numbers appearing on them is 25?

  1. 0
  2. 1/2
  3. 1
  4. 1/1296

Answer (Detailed Solution Below)

Option 1 : 0

Probability of Random Experiments Question 13 Detailed Solution

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Concept:

Probability of an event happening = 

If a die thrown, Number of sample space = 6, If two dice are thrown n(S) = 62 = 36

Calculation:

Here, four dice are thrown, 

n(S) = 64

Now, sum of the numbers appearing on them 25 = { }       

⇒ n = 0               

(∵maximum sum = 6 + 6 + 6 + 6 = 24)

∴ Probability = 0/(64) = 0

Hence, option (1) is correct.

From a pack of playing card, one card is drawn randomly. What is the probability that the card is red color or king?

Answer (Detailed Solution Below)

Option 4 :

Probability of Random Experiments Question 14 Detailed Solution

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Concept:

  • Either event A alone OR event B alone: m + n.
  • Both event A AND event B together: m × n.

 

Calculation:

There are 26 red cards out of total of 52 cards which also include 2 kings 

So the probability of getting a red card (P1) = 

Now from 4 kings as 2 kings are already counted there 2 kings are left

So the probability of getting either of them (P2) = 

∴ The probability that the card is red colour or king (P) = P1 + P2

P = 

P = 

A coin is tossed 3 times. The probability of getting a head and a tail alternately is:

Answer (Detailed Solution Below)

Option 1 :

Probability of Random Experiments Question 15 Detailed Solution

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Concept:

The probability of the occurrence of an event A, out of total possible outcomes N, is given by P(A) = , where n(A) is the number of ways in which the event A can occur.

 

Calculation:

The total number of different possible outcomes (N) in tossing a coin 3 times is 23 = 8.

For getting a head and a tail alternately, the possibilities are HTH, THT → 2 possibilities n(A).

∴ Required probability = 

Alternate Method

The possible set of A coin is tossed 3 times is {HHH}{HHT}{HTH}{HTT}{TTT}{TTH}{THT}{THH} = 8

The probability of getting a head and a tail alternately is {HTH}{THT} = 2

So, required probability = 

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