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

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

Calculation:

Given,

Mean of the binomial distribution,

Variance of the binomial distribution,

For a binomial random variable,

From the mean,

Using the variance,

Simplifying,

∴ The number of trials is .

Calculation:

Given,

Probability of hitting the target per shot,

Number of shots fired,

The number of hits follows a binomial distribution:

Probability of zero hits:

Probability of exactly one hit:

Probability of at least two hits:

∴ The probability of hitting the target at least twice is  .

Hence, the correct answer is Option 3. 

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.

Calculation:

Given,

We need to calculate, the probability that neither A  nor B  occurs.

Using the complement rule:

Now, apply the inclusion-exclusion formula to find :

Substitute the given values:

To simplify:

Now, calculate :

Hence, the correct answer is Option 2.

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 =  +  = 

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 

Concept:

• For two events A and B, we have: P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
• If A and B are independent events, then P(A ∩ B) = P(A) × P(B).

Calculation:

Using the concept above, because A and B are independent events, we can write:

P(A ∪ B) = P(A) + P(B) - P(A) × P(B)

⇒ 0.7 = 0.4 + P - 0.4 × P

⇒ 0.6P =0.3

⇒ P = 0.5.

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 = 

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 =

Concept:

• A ⊂ B = Proper Subset: every element of A is in B, but B has more elements.
• ϕ = Empty set = {}

Calculation:

Given: P(B/A) = 1

⇒ 

⇒ P(A ∩ B) = P(A)

⇒ (A ∩ B) = A

So, every element of A is in B, but B has more elements.

∴ A ⊂ B

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 = 

Concept:

Permutations with Repetition = nr

Where n is the number of things to choose from r different things when repetition is allowed, and order matters.

Favorable cases = Total cases - Unfavorable cases

Calculation:

According to the question

Four dies are rolled 

So, Total Possible number of outcomes = 64

Now, Total outcomes when no 2 appears = 54

Now, From the concept used

Favorable cases = 64 - 54

⇒ 1296 - 625

⇒  671

∴ The number of possible outcomes in which at least one die shows 2 is 671.

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}

= 

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