Test Level 1: Probability - 1 (September 30) - CAT MCQ

Favourable cases = 3, 6, 9, 12, 15, 18, 7, 14
Probability = 8/20 = 2/5

Probability that the work is done

Number of Ace except ace of spade = 4 - 1 = 3
Number of spade except ace of spade = 13 - 1 = 12

S = sample space.
n(S) = 36.
E = event of getting a prime number = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}.
n(E) = 15.
P(not a prime number) = (1 - (15/36)) = 21/36 = 7/12.

Total cases = 63
Favorable cases = 43
Probability of getting a number less than 5 in each case

5 coins can be selected at random as 2 + 4 + 6 = 12 coins in C(12, 5) ways.
They will amount to at least Rs. 1.50 if two 50 paisa and three 25 paisa coins are selected or one 50 paisa and four 25 paisa coins are selected or two 50 paisa coins, two 25 paisa coins and one 10 paisa coin are selected.
This can be done in ²C₂ × ⁴C₃ + ²C₁ × ⁴C₄ + ²C₂ × ⁴C₂ × ⁶C₁ = (4 + 2 + 36) = 42 ways.
Thus, required probability is

Favourable cases = 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Number of favourable cases = 10
Required probability = 10/100 = 1/10

P(win) = 4/36 = 1/9
P(loss) = 8/9

Let the value of m = 2 and n = 1
Total outcomes will be 8, and probability of getting two consecutive outcomes will be
HHT, HHH and THH, i.e.3/8, only option 3 satisfies for m = 2 and n = 1