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

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answer:

i havent taken the psat, but i have the sat, make sure to bring some snacks, because youre in there forever, 2 sharp number two pencils, an sat approved calculator, and maybe a jacket.   your phone and any books will be taken up, so don't try to bring them(but they will be given back at the end of the testing).   good luck!   i hope you do well!