# 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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To find the probability that exactly 5 out of 20 parts are too large, you can use the binomial probability formula:

[ P(X = k) = \binom{n}{k} \times p^k \times (1 - p)^{n - k} ]

Where:

- ( P(X = k) ) is the probability of getting exactly ( k ) successes,
- ( n ) is the number of trials (in this case, 20 parts),
- ( k ) is the number of successes (in this case, 5 parts being too large),
- ( p ) is the probability of success on each trial (in this case, 15% or 0.15).

Plug in the values:

[ P(X = 5) = \binom{20}{5} \times 0.15^5 \times (1 - 0.15)^{20 - 5} ]

Calculate:

[ P(X = 5) = \binom{20}{5} \times 0.15^5 \times 0.85^{15} ]

[ P(X = 5) = \frac{20!}{5!(20-5)!} \times 0.15^5 \times 0.85^{15} ]

[ P(X = 5) = \frac{20!}{5!15!} \times 0.15^5 \times 0.85^{15} ]

[ P(X = 5) ≈ 0.226 ]

So, the probability that exactly 5 out of 20 parts are too large is approximately 0.226 or 22.6%.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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