Questions: Question 30 2.86 pts Find the indicated probability. A bin contains 77 light bulbs of which 10 are defective. If 6 light bulbs are randomly selected from the bin with replacement, find the probability that all the bulbs selected are good ones. Round to the nearest thousandth if necessary. 0.567 0.434 0.848 0.664

Solution Steps

The total number of light bulbs in the bin is \( 77 \), and the number of defective bulbs is \( 10 \). Therefore, the number of good bulbs is given by:

\[ \text{Good bulbs} = 77 - 10 = 67 \]

The probability \( p \) of selecting a good bulb is calculated as:

\[ p = \frac{\text{Good bulbs}}{\text{Total bulbs}} = \frac{67}{77} \]

We are selecting \( n = 6 \) light bulbs with replacement. We want to find the probability that all selected bulbs are good, which means we are looking for \( k = 6 \) successes (good bulbs).

The probability of exactly \( k \) successes in \( n \) trials for a binomial distribution is given by the formula:

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

In our case, we need to calculate:

\[ P(X = 6) = \binom{6}{6} \left(\frac{67}{77}\right)^6 \left(1 - \frac{67}{77}\right)^{6-6} \]

Since \( \binom{6}{6} = 1 \) and \( (1 - p)^{0} = 1 \), the formula simplifies to:

\[ P(X = 6) = \left(\frac{67}{77}\right)^6 \]

After performing the calculations, we find that the probability that all selected bulbs are good is:

\[ P(X = 6) \approx 0.434 \]

This value represents the probability that all 6 light bulbs selected from the bin are good ones.

Final Answer