The problem involves selecting 5 keyboards randomly from a group of 25, with 19 having mechanical defects and 6 having electrical defects. We are tasked with calculating the probability that at least 4 of the 5 selected keyboards have mechanical defects.

Approach:

This is a combinatorial probability problem where we will use the hypergeometric distribution to calculate the probability. The hypergeometric formula is:

$P(X = k) = \frac{\binom{M}{k} \binom{N-M}{n-k}}{\binom{N}{n}}$

Where:

• $N$ = total population (25 keyboards),
• $M$ = number of mechanical defect keyboards (19),
• $n$ = sample size (5 keyboards),
• $k$ = number of keyboards with mechanical defects in the sample.

We want to find the probability that at least 4 out of 5 keyboards have mechanical defects. Thus, we need to calculate:

$P(\text{at least 4 mechanical defects}) = P(X = 4) + P(X = 5)$

Step-by-step Calculation:

1. Probability of exactly 4 mechanical defects ($P(X = 4)$): $P(X = 4) = \frac{\binom{19}{4} \binom{6}{1}}{\binom{25}{5}}$
2. Probability of exactly 5 mechanical defects ($P(X = 5)$): $P(X = 5) = \frac{\binom{19}{5} \binom{6}{0}}{\binom{25}{5}}$

Let's calculate these values.The probabilities are as follows:

• The probability of selecting exactly 4 keyboards with mechanical defects, $P(X = 4)$, is approximately 0.438.
• The probability of selecting exactly 5 keyboards with mechanical defects, $P(X = 5)$, is approximately 0.219.
• Therefore, the total probability that at least 4 keyboards have mechanical defects, $P(\text{at least 4})$, is approximately 0.657.

Would you like more details or have any further questions?

Here are 5 related questions for practice:

1. What is the probability that exactly 3 keyboards have mechanical defects?
2. How would the probability change if we selected 6 keyboards instead of 5?
3. Can you calculate the expected number of keyboards with mechanical defects in a sample of 5?
4. How would you compute the probability if the number of keyboards with mechanical defects was 15 instead of 19?
5. How would the probability change if we wanted at least 3 keyboards with mechanical defects?

Tip: The hypergeometric distribution is useful when sampling without replacement, as in this case where the population size is limited.