Question

Three fair coins are tossed simultaneously 10 times. Find the probability that "2 heads and one tail" will show up (a) at least once and (b) at most once.

Answer 1

If we will toss the three fair coin simultaneously only once, the the probability that "2 heads and one tail" will show up is = 3/8

So, here

p = 3/8 = 0.375

n = 10

It is a binomial distribution.

P(X = x) = nCx * p^x * (1 - p)^{n - x}

a) P(X > 1) = 1 - P(X < 1)

                 = 1 - P(X = 0)

                = 1 - (10C0 * (0.375)^0 * (0.625)^10)

               = 1 - 0.0091 = 0.9909

b) P(X < 1) = P(X = 0) + P(X = 1)

                  = 10C0 * (0.375)^0 * (0.625)^10 + 10C1 * (0.375)^1 * (0.625)^9

                  = 0.0637