Products from a certain machine are too large 15% of the time. What is the probability that in a run of 20 parts, 5 are too large?

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Products from a certain machine are too large 15% of the time. What is the probability that in a run of 20 parts, 5 are too large?

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Answer 1 · 2018-06-25T15:46:23

$C_{5}^{20} {(.15)}^{5} {(.85)}^{15} \approx .103$ $C_5^20(.15)^5(.85)^15~~.103$

Explanation:

The probability that 5 parts are too large is $.15 \times .15 \times .15 \times .15 \times .15 = {(.15)}^{5}$ $.15xx.15xx.15xx.15xx.15=(.15)^5$ and since 5 parts are too large, the other 15 are of acceptable size with probability ${(1 - .15)}^{15}$ $(1-.15)^15$ . Since any 5 parts can be too large in no particular order, there are $C_{5}^{20}$ $C_5^20$ ways to do this, so the probability that exactly 5 parts are too large is $C_{5}^{20} {(.15)}^{5} {(.85)}^{15} \approx .103$ $C_5^20(.15)^5(.85)^15~~.103$ .

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Products from a certain machine are too large 15% of the time. What is the probability that in a run of 20 parts, 5 are too large?

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Explanation:

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