There are 8 cars in a parking lot on a very cold day. Suppose the probability of any of them not starting is 0.13. What is the probability that exactly 2 of the cars will not start?

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There are 8 cars in a parking lot on a very cold day. Suppose the probability of any of them not starting is 0.13. What is the probability that exactly 2 of the cars will not start?

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Answer 1 · 2018-05-15T01:06:38

I don't have a CAS on me right now but you can use the explanation to find the fraction form
Answer: 0.20519191831

Explanation:

Use binomial distribution formula:

$8 C 2 {(0.13)}^{2} {(1 - 0.13)}^{8 - 2}$ $8C2(0.13)^2(1-0.13)^(8-2)$

Basically it's:
how many combinations are there for out of 8 cars, 2 won't start
$\times$ $xx$
probability that two cars won't start ${(0.13)}^{2}$ $(0.13)^2$
$\times$ $xx$
probability that the rest will start ${(1 - 0.13)}^{8 - 2}$ $(1-0.13)^(8-2)$

To solve it use the combination formula:

$\frac{8!}{(8 - 2)! 2!} \times {(0.13)}^{2} {(1 - 0.13)}^{8 - 2}$ $(8!)/((8-2)!2!)xx(0.13)^2(1-0.13)^(8-2)$

$= 28 \times 0.0169 \times 0.433626201 = 0.20519191831$ $=28xx0.0169xx0.433626201=0.20519191831$

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There are 8 cars in a parking lot on a very cold day. Suppose the probability of any of them not starting is 0.13. What is the probability that exactly 2 of the cars will not start?

1 Answer

Explanation:

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