The probability that the space shuttle is launched on the designated day is 80%. Assume that
1 answer:
Answer:
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
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