The probability that the space shuttle is launched on the designated day is 80%. Assume that shuttle launches are independent from each other. Suppose four launches are scheduled in the
next three months. What is the probability that more than one is launched on the designated
day?

0.9728

Explanation:

P(launch), p = 80% = 0.8

(1 - p) = 1 - 0.80 = 0.2

Number of launches = 4

We can apply the binomial probability formula :

P(x =x) = nCx * p^x * (1 - p)^(n - r)

Probability that More than 1 is launched

P(x > 1) = p(x = 2) + p(x = 3) + p(x = 4)

P(x = 2) :

4C2 * 0.8^2 * 0.2^2

6 * 0.64 * 0.04 = 0.1536

P(x = 3) :

4C3 * 0.8^3 * 0.2^1

4 * 0.512 * 0.2 = 0.4096

P(x = 4) :

4C4 * 0.8^4 * 0.2^0

1 * 0.4096 * 1 = 0.4096

P(x > 1) = 0.1536 + 0.4096 + 0.4096

P(x > 1) = 0.9728

she has insufficient time to complete her important tasks

hope this : )