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This is a probability problem involving jury selection. The problem asks you to find probabilities associated with choosing jurors from a pool of 16 men and 12 women, selecting 12 jurors. Let me guide you through the problem step by step.

### Definitions and notation:
- \( M = 16 \): Number of men in the pool.
- \( W = 12 \): Number of women in the pool.
- \( n = 12 \): Number of jurors to be selected.

We use **combinatorics** for calculating probabilities based on combinations (denoted by \( \binom{n}{k} \), which is "n choose k").

### Part (a): **Probability that the jury is made up of 2 men.**
To find the probability that exactly 2 men are chosen, you can use the formula for combinations:
\[
P(\text{2 men}) = \frac{\binom{16}{2} \times \binom{12}{10}}{\binom{28}{12}}
\]
- \( \binom{16}{2} \): Choosing 2 men from 16.
- \( \binom{12}{10} \): Choosing 10 women from 12.
- \( \binom{28}{12} \): Total number of ways to choose 12 jurors from 28 people.

### Part (b): **Probability that half of the jury are men and half are women.**
Here, you want to find the probability that 6 men and 6 women are selected:
\[
P(\text{6 men, 6 women}) = \frac{\binom{16}{6} \times \binom{12}{6}}{\binom{28}{12}}
\]
- \( \binom{16}{6} \): Choosing 6 men from 16.
- \( \binom{12}{6} \): Choosing 6 women from 12.

### Part (c): **Probability that there are at least 2 women on the jury.**
This is the complement of the event that fewer than 2 women are selected, i.e., the probability of 0 or 1 woman. So:
\[
P(\text{at least 2 women}) = 1 - P(\text{0 women or 1 woman})
\]
First, calculate the probabilities for 0 and 1 woman:
\[
P(\text{0 women}) = \frac{\binom{16}{12} \times \binom{12}{0}}{\binom{28}{12}}, \quad P(\text{1 woman}) = \frac{\binom{16}{11} \times \binom{12}{1}}{\binom{28}{12}}
\]
Then subtract from 1:
\[
P(\text{at least 2 women}) = 1 - (P(\text{0 women}) + P(\text{1 woman}))
\]

### Part (d): **Probability that 9 men are selected.**
Here, you're asked for the probability of selecting exactly 9 men:
\[
P(\text{9 men}) = \frac{\binom{16}{9} \times \binom{12}{3}}{\binom{28}{12}}
\]
- \( \binom{16}{9} \): Choosing 9 men from 16.
- \( \binom{12}{3} \): Choosing 3 women from 12.

### Part (e): **Analysis of random jury selection and suspicion.**
This part asks for an interpretation of whether selecting 9 men (as found in part d) is likely or unlikely under random selection. Based on the probability calculated in part d, if the chance is very small, it might raise suspicion about whether the selection was random.

#### Do you want me to calculate these values and clarify any part of the process?

A pool of potential jurors consists of 16 men and 12 women. 12 jurors are to be chosen at random from the pool of 28. Find the probabilities for different jury compositions (2 men, half men/half women, at least 2 women, exactly 9 men), and determine if a jury of 9 men is suspicious.

Explore the probabilities associated with selecting a jury of 12 people from a pool of 16 men and 12 women. This problem involves combinatorics and probability calculations, including the probability of choosing 2 men, half men and half women, and the likelihood of selecting at least 2 women.

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