Total number of caps = 12
n(s) = ¹²c₁ = 12
Out of ( 2 blue + 1 yellow) caps, number of ways to pick one cap n(E) = ³c₁ = 3
∴ Required probability = p(E) = n (E) / n(s) = 3/12 = 1/4

Correct Option: B

Total number of caps = 12
n(s) = ¹²c₁ = 12
Out of ( 2 blue + 1 yellow) caps, number of ways to pick one cap n(E) = ³c₁ = 3
∴ Required probability = p(E) = n (E) / n(s) = 3/12 = 1/4

Total number of ways of selection of 3 marbles out of 12
= n(S) = ¹²C₃ = 220
Total number of favorable events = n(E)
= ³C₃ + ⁴C₃ = 1 + 4 = 5

Correct Option: D

Total number of ways of selection of 3 marbles out of 12
= n(S) = ¹²C₃ = 220
Total number of favorable events = n(E)
= ³C₃ + ⁴C₃ = 1 + 4 = 5
∴ Required probability
= 5/220 = 1/44

Required probability = 1- probability that all the girls sit together

Correct Option: A

Required probability = 1- probability that all the girls sit together
= 1 - 1/42 = 41/42

Required probability = 1- probability that all the boys sit together

Correct Option: C

Required probability = 1- probability that all the boys sit together
= 1- 1/42 = 41/42

We have, 5 boys and 5 girls.
Since, all the girls are sitting together, we consider them as one. Now, we can arrange 5 boys and 1 girl in 6! ways and these 5 girls (whom we considered as one) can also be arranged in 5! ways.
∴ Favorable number of ways = 6!5!
and total number of ways to arrange 5 boys and 5 girls = 10!

Correct Option: A

We have, 5 boys and 5 girls.
Since, all the girls are sitting together, we consider them as one. Now, we can arrange 5 boys and 1 girl in 6! ways and these 5 girls (whom we considered as one) can also be arranged in 5! ways.
∴ Favorable number of ways = 6!5!
and total number of ways to arrange 5 boys and 5 girls = 10!
∴ Required probability 6!5!/10! = 1/42