Daily Practice Questions Questions for CAT with Answers PDF

Probability of selecting two red balls = 
Odds in favour = 3 : 7
Odds against = 7 : 3

S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH }.
E = {HHT, HTH, THH, HTT, THT, TTH, HHH}.
Therefore, P(E) = 7/8

Total outcomes = 36
Favourable outcomes for getting a doublet or a total of 10 = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (4, 6), (6, 4)]
Probability = 8/36
Probability that neither a doublet nor a total of 10 will appear 

Hence, answer option 1 is correct.

There are ¹⁰C₄ × 4! = 5040 sequences of 4 distinct digits, out of which there is only one sequence in which the lock opens.
Required probability = 1/5040

Total number of possible cases = 6³ = 216
Favourable number of cases = 6
Probability = 6/216 = 1/36

Number of ways of choosing 1 girl out of 10 = ¹⁰C₁
Number of ways of choosing 2 boys out of 15 = ¹⁵C₂
Total number of ways of choosing 3 students out of 25 (15 + 10) = ²⁵C₃

Total cases = 6² 
Both odd numbers
 

Number of favourable outcomes = 300, as heads has appeared 300 times.
Total number of possible outcomes = 700, as the coin is tossed 700 times.
⇒ Probability = 300/700 = 3/7

Probability that at least one of A and B will solve the problem

P(None is defective)

P(at least one is defective) 

Probability that at least one horse will win = 1 - Probability that neither horse will win

S = Sample space = 36.
E = Event of getting the sum as 6.
Favorable outcomes = {(2,4), (4,2), (1,5), (5,1), (3,3)}.
So, n(E) = 5.
P(E) = 5/36.

Total number of cases when a coin is flipped five times = 32
Favourable number of cases (2 heads and 3 tails) = ⁵C₂ or ⁵C₃ = 10
Required probability = 10/32 = 5/16

Total number of outcomes = ¹⁵C₃
Number of favourable outcomes = ⁹C₃

E = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2), (6, 4)}
n(E) = 11
n(s) = 36

Total number of sample spaces = 7!
Total number of favourable outcomes = 3! × 5!
Required probability = (3! × 5!)/7!
= 1/7

Let A be the event that at least two people have the same birthday.
Then,

Out of a total of 6 occurrences, 3 is one possibility = 1/6.

Event definition is:
(1 and 1) or (1 and 2) or (1 and 3) or (1 and 4) or  
(1 and 5) or (1 and 6) or (2 and 1) or (3 and 1) or
(4 and 1) or (5 and 1) or (6 and 1)
Total 11 out of 36 possibilities = 11/36

Knave and queen or Queen and Knave → 4/52 × 4/51 + 4/52 × 4/51 = 8/663