If three marbles are picked at random, what is the probability that either all are green or all are red?

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

Required probability = favorable number of cases/ total number of
=⁸P₅ / 8 ⁵ = 105/512

Total number of possible arrangement for 4 boy and 3 girls in a queue = 7!
when they occupy alternate position then the arrangement would be like
BGBGBGB. Thus the total number of possible arrangement = 4! x 3!
? Required probability = 4! x 3! / 7!
= (4 x 3 x 2 x 3 x 2) / (7 x 6 x 5 x 4 x 3 x 2) = 1/35

First we find the probability that the plane is not hit
= (1 - 0.4) ( 1 - 0.3) (1 - 0.2) (1 - 0.1)
= 0.6 x 0.7 x 0.8 x 0.9 = 0.3024

? probability that the plane is hit = 1 - 0.3024
= 0.6976

Since, there are only two types of socks in the bag. so, if murari pick up 3 socks then certainly two of them are of same type. Thus this is a certain event.
Hence required probability = 1

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
= 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!
? Required probability 6!5!/10! = 1/42

Required probability = 6!5!/10! = 1/42

n(S) = 2¹⁰⁰
n(E) = No. of favourable ways = ¹⁰⁰C₁ + ¹⁰⁰C₃ + ... ¹⁰⁰ C₉₉ = 2^100-1 = 2⁹⁹
[? ⁿC₁ + ⁿC₃ + ⁿC₅ + ........................... = 2^n-1 ]
? P(E)= n(E)/n(S) = 2⁹⁹/2¹⁰⁰ = 1/2

Note : The given case can be generalised as "If a unbiased coin is tossed 'n' times, then the chance that the head will present itself an odd number of times is 1/2. "