Daily Practice Questions Questions for CAT with Answers PDF

E1 = Getting 5 or 6 in a single throw of a die.
E2 = Getting 1, 2, 3 or 4 in a single throw of a die.
A = Getting exactly one head.
P(E1) = 2/6 = 1/3.
P(E2) = 4/6 = 2/3.
P(A/E1) = Probability of getting exactly one head when a coin is tossed three times = ³C₁ × (1/2)¹ × (1/2)² = 3/8.
P(A/E2) = Probability of getting exactly one head when a coin is tossed once only = 1/2.
Now, Required probability = P(E2/A) 

Probability that the 1st ball is white and the 2nd ball is other than white =

Probability that the 1st ball is red and the 2nd ball is other than red 

Probability that the 1st ball is green and the 2nd ball is other than green

Probability that the 1st ball is black and the 2nd ball is other than black 

Probability that both the balls are of different colours 

Probabilities that 3 men hit a target are 1/6 , 1/4 and 1/3 and
Probabilities that the same 3 men do not hit the target are 5/6, 3/4 and 2/3.
Let,
Event (A) = Exactly one of them hits the target 1st person or 2nd person or 3rd person

The possible cases of taking out 2 red and 1 blue balls are:

Let Z₁ = Event that the shirt is made by tailor A;
Z₂ = Event that the shirt is made by tailor B;
Z₃ = Event that the shirt is made by tailor C; and
z = Event that the shirt is defective.        
Total number of shirts made by these three tailors = 500 + 1200 + 1600 = 3300
P(Z₁) = Probability that the shirt is made by tailor A = 500/3300 = 5/33
P(Z₂) = Probability that the shirt is made by tailor B = 1200/3300 = 4/11
P(Z₃) = Probability that the shirt is made by tailor C = 1600/3300 = 16/33
Total number of defective shirts made by three tailors = 200 + 300 + 300 = 800
Given, P(z/Z₁) = Probability that the shirt is defective given that it is made by tailor A = 200/800 = 1/4 = 0.25
Similarly, P(z/Z₂) = 300/800 = 3/8 = 0.375 and P(z/Z₃) = 0.375
Required probability = Probability that the shirt is made by tailor C given that the shirt is defective = P(Z₃/Z)



Required probability = 600/1175 = 24/47
{Baye's Theorem is given by P(Ei/A) 

Total number of possible ways = 3!
Number of favourable cases = 1
Thus, required probability = 1/3! = 1/6

One possibility is that all three of them hit the 1st target and miss out on the 2nd target.
The probability for this is 0.95 × 0.30 × 0.85 × 0.20 × 0.90 × 0.25 = 0.01090125
The 2nd possibility is that all of them hit only the 2nd target and miss out on the 1st.
The probability for the same is 0.05 × 0.7 × 0.15 × 0.80 × 0.10 × 0.75 = 0.000315
Thus, required probability = 0.01090125 + 0.000315 = 0.01121625 ~ 0.011

A white ball can be drawn in two mutually exclusive ways.
(i) Selecting bag X and then drawing a white ball from it.
(ii) Selecting bag Y and then drawing a white ball from it.
Let E1: event of selecting bag X
E2: event of selecting bag Y
A: event of drawing white ball
∴ P(E1) = P(E2) = 1/2
Now, P(A/E₁) = Probability of drawing a white ball when bag X is chosen = 3/5
P(A/E₂) = Probability of drawing a white ball when bag Y is chosen = 2/6
P(white ball) = P(A) = P(E₁) P(A/E₁) + P(E₂) P(A/E₂) (Law of total probability)

There are 3 favourable cases for the same which shall be discussed one by one.
The 1st case is in which A scores a 1 in both the rolls and B does not get to roll the die.
The probability for the same is (1/2)(1/6)(1/2)(1/6)(1/2)(1/2) = 1/576
The 2nd case is in which A gets a score of 2 in his first roll, gets no score in the 2nd roll and B gets no score in either of his rolls.
The probability for the same is (1/2)(1/6)(1/2)(1/2)(1/2) = 1/96
The 3rd case is in which A gets a score of 0 in the 1st roll, gets a score of 2 in the 2nd roll and B gets no score in either of his rolls.
The probability for the same is (1/2)(1/2)(1/6)(1/2)(1/2) = 1/96
Thus, the required probability is 1/96 + 1/96 + 1/576 = 13/576

There shall be two cases for the same.
Case I: A scores an even number in his first roll and an odd number in his second roll, the probability of which is (1/2)(3/6)(1/2)(3/6) = 1/16
Case II: A scores an odd number in his first roll and an even number in his second roll, the probability of which is (1/2)(3/6)(1/2)(3/6) = 1/16
Thus, the required probability is 1/16 + 1/16 = 1/8