Probability Test

Result

Probability Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

A number is chosen randomly from the set {1, 2, 3, …15}. If chosen number is a two digit number, then the probability that the number is a prime numbers, is

A
1/2

B
1/3

C
1/4

D
1/5

Solution

Required probability = 2 / 6 = 1/3
Hence, option B is correct.
Question 2
1 / -0

A single letter is selected at random from the word 'PROBABILITY'. The probability that it is a vowel is –

A
3/11

B
2/11

C
4/11

D
7/11

Solution

PROBABILITY
vowel = O, A, I, I
Total case = ¹¹C₁ = 11
Favourable case = ⁴C₁ = 4
Required probability = 4/11
Question 3
1 / -0

Let A, B, C be three mutually independent events. Consider the two statements S₁ and S₂
S₁ : A and B ∪ C are independent
S₂ : A and B ∩ C are independent
Then,

A
Both S₁ and S₂ are true

B
Only S₁ is true

C
Only S₂ is true

D
Neither S₁ nor S₂ is true

Solution

We are given that

P(A ∩ B) = P(A) P(B)
P(B ∩ C) = P(B) P(C), P(C ∩ A) = P(C) P(A)
And P( A ∩ B ∩ C)=P(A) P(B) P(C)
=P(A) P(B) P(C) = P(A) P(B ∩ C).
⟹ A and B ∩ C are independent. Therefore, S2 is true.
Also P[(A ∩ (B ∪ C)]= P[(A ∩ B) ∪ (A ∩ C)]
=P(A ∩B) + P(A ∩ C) –P[A ∩ B) ∩ (A ∩ C)]
=P(A ∩ B) + P(A ∩ C) – P(A ∩ B ∩ C)

=P(A) P(B) +P(A) P(C) – P(A) P(B) P(C)
=P(A) [P(B) +P(C)-P(B) P(C)]
=P(A) [P(B) + P(C) –P(B ∩ C)] = P(A) P(B ∪ C)
∴ A and B ∪ C are independent.