A. 25/42
B. 9/20
C. 10/23
D. 41/42
Explanation:
Total fruits = 9
Since there must be at least two apples,
(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42.

A. 8/15
B. 2/5
C. 3/5
D. 7/15
Explanation:
Drawing two balls of same color from seven green balls can be done in ⁷C₂ ways.
Similarly from eight white balls two can be drawn in ⁸C₂ ways.
P = ⁷C₂/¹⁵C₂ + ⁸C₂/¹⁵C₂ = 7/15

A. 1/20
B. 3/10
C. 1/5
D. 4/5
Explanation:
Six persons can be arranged in a row in 6! ways. Treat the three persons to sit together as one unit then there four persons and they can be arranged in 4! ways. Again three persons can be arranged among them selves in 3! ways. Favourable outcomes = 3!4! Required probability = 3!4!/6! = 1/5

A. 19/52
B. 17/52
C. 5/13
D. 4/13
Explanation:
P(SᴜK) = P(S) + P(K) – P(S∩K), where S denotes spade and K denotes king.
P(SᴜK) = 13/52 + 4/52 – 1/52 = 4/13

A. 3/8
B. 6/11
C. 2/5
D. 4/5
Explanation:
Since four coins are tossed, sample space = 24
Getting two heads and two tails can happen in six ways.
n(E) = six ways
p(E) = 6/24 = 3/8

A. 1/4
B. 1/2
C. 3/4
D. 3/5
The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 – 1/2 = 1/2.

A. 1/2
B. 16/19
C. 4/5
D. 17/20
Explanation:
n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
P(EᴜP) = 10/20 + 8/20 – 1/20 = 17/20

A. 1/25
B. 1/2
C. 4/13
D. 1/10
Explanation:
We have 1, 8, 27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is
4/100 = 1/25.

A. 31/32
B. 1/16
C. 1/2
D. 1/32
Explanation:
Let P(T) be the probability of getting least one tail when the coin is tossed five times.
= There is not even a single tail.
i.e. all the outcomes are heads.
= 1/32 ; P(T) = 1 – 1/32 = 31/32

A. 21/25
B. 17/25
C. 4/25
D. 8/25
Explanation:
The number of exhaustive events = ⁵⁰C₁ = 50.
We have 15 primes from 1 to 50.
Number of favourable cases are 34.
Required probability = 34/50 = 17/25.