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.