The image you've uploaded contains several problems related to probability distributions. I'll help you work through each problem one by one. Let's begin with Problem 5.53, which involves defective computer chips.

Problem 5.53 Defective Computer Chips

We are given six computer chips, three of which are defective. Three chips are randomly chosen for inspection, and we are asked to find the probability distribution for the number of defective chips in the sample of three chosen.

Step 1: Defining the Random Variable

Let $X$ be the random variable representing the number of defective chips in the sample of 3.

• The total number of computer chips is 6, with 3 defective ones.
• We are choosing 3 chips randomly.

Step 2: Possible Values of $X$

The possible values of $X$ (the number of defective chips chosen) can be 0, 1, 2, or 3.

Step 3: Computing the Probability for Each Value of $X$

The probability of each outcome can be computed using combinations, since this is a hypergeometric distribution (since we are sampling without replacement). The general formula for a hypergeometric probability is:

$P(X = k) = \frac{\binom{3}{k} \binom{3}{3-k}}{\binom{6}{3}}$ where:

• $\binom{3}{k}$ is the number of ways to choose $k$ defective chips from 3 defective chips.
• $\binom{3}{3-k}$ is the number of ways to choose the remaining chips from the 3 non-defective chips.
• $\binom{6}{3}$ is the total number of ways to choose 3 chips from 6 chips.

Step 4: Calculating for Each $k$

• For $X = 0$ (no defective chips): $P(X = 0) = \frac{\binom{3}{0} \binom{3}{3}}{\binom{6}{3}} = \frac{1 \times 1}{20} = \frac{1}{20}$

• For $X = 1$ (one defective chip): $P(X = 1) = \frac{\binom{3}{1} \binom{3}{2}}{\binom{6}{3}} = \frac{3 \times 3}{20} = \frac{9}{20}$

• For $X = 2$ (two defective chips): $P(X = 2) = \frac{\binom{3}{2} \binom{3}{1}}{\binom{6}{3}} = \frac{3 \times 3}{20} = \frac{9}{20}$

• For $X = 3$ (three defective chips): $P(X = 3) = \frac{\binom{3}{3} \binom{3}{0}}{\binom{6}{3}} = \frac{1 \times 1}{20} = \frac{1}{20}$

Step 5: Final Probability Distribution

The probability distribution for the number of defective chips $X$ is:

$P(X = 0) = \frac{1}{20}, \quad P(X = 1) = \frac{9}{20}, \quad P(X = 2) = \frac{9}{20}, \quad P(X = 3) = \frac{1}{20}$

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Next Steps

Would you like me to continue solving the next problems (5.54 and 5.55)? Or do you have any questions on this part?