Test: Engineering Mathematics- 7 - Computer Science Engineering (CSE) MCQ

Since A and B are exhaustive events

∴ P(A∪B) = 1

Let the number of students be 100.

P(N)→ probability that students failed in Computer Networks

P(A)→ probability that students failed in Computer Architecture

P(N∩A)→ probability that the student has failed in both the subject

 

P(NUA)→ probability that the  student has failed in atleast one subject

P(NUA) = P(N) + P(A) − P(N∩A) 

Required probability = 

Let the events be defined as:

A: Obtaining a sum of 8

B: Getting an even number on both dice

P(A U B) = P(A) + P(B) – P (A ∩ B)

Now cases favourable to A are (3, 5) (5, 3) (2, 6) (6, 2) (4, 4)

So, P(A) = 5/36

Cases favourable to B: (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 2), (6, 4), (6, 6).

P(B) = 9/36

Now, (2, 6) (6, 2) and (4, 4) are common to both events A and B

So, P (A ∩ B) = 3/36

⇒ P (A ∪ B) = (5/36) + (9/36) – (3/36) = 11/36

P(A)→ probability of getting a blue ball from the 2^nd bag.

P(E1)→ probability of getting a red ball from the 1^st bag.

P(E2)→ probability of getting a blue ball from the 1^st bag.

P(X)→ probability that bonus scheme is introduced

P(A) → probability of A being a manager.

P(B) → probability of B being a manager.

P(C) → probability of C being a manager.

P(X) = P (A ∩ X) + P (B ∩ X) + P (C ∩ X)

P(X) = P(A)P(X/A) + P(B)P(X/B) + P(C)P(X/C)

T: A speaks truth ⇒ P(T) = ¾

T̅ : A lies ⇒ P(T̅) = 1 - P(T) = ¼
 

There are three cases for matches. It can be won, drawn or lost.

The probability of winning a match, P (B/T) = 1/3

The probability of not winning a match, P(B,T̅) = 2/3

Using Baye’s theorem:

The sum 5 can be obtained as follows:

(1, 4), (2, 3), (3, 2), (4, 1)

The probability of A is throwing 5 with two dice is

= 4/36 = 1/9

The probability of A is not throwing 5 is

 

The sum 6 can be obtained as follows:

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

The probability of B is throwing 6 is

 = 5/36

The probability of B is not throwing 6 is

Given that A begins,

Now A can win if he throws 5, in first chance,

(or) third Chace (b) fifth chance etc –

The chances of winning of A

Let p₁ = 0.4, p₂ = 0.3, p₃ = 0.2 and p₄ = 0.1

P(the gun hits the plane) = P(the plane is hit at least once)

= 1 – P(the plane is hit in none of the shots)

= 1− (1−p₁) (1−p₂) (1−p₃) (1−p₄)
= 1− (0.6 × 0.7 × 0.8 × 0.9)
= (1− 0.3024) = 0.6976

P(A)→ probability of getting more than one tail

∴ P(A) = 11/16

P(B)→ probability of getting at least one tail

∴ P(B) = 15/16

P(A ∩ B)→ probability of getting more than one tail and at least one is head.

∴ P(A ∩ B) = 10/16