This is a binomial probability problem, where we want to find the probability of exactly 3 left-handed people out of 5 randomly selected individuals.

The binomial probability formula is:

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

Where:

• $n = 5$ (the total number of trials, i.e., the number of people selected),
• $k = 3$ (the number of successful outcomes, i.e., 3 left-handed people),
• $p = 0.13$ (the probability of success on a single trial, i.e., being left-handed),
• $\binom{n}{k}$ is the binomial coefficient, representing the number of ways to choose 3 lefties out of 5 people.

Step-by-Step Calculation:

1. Binomial coefficient $\binom{n}{k}$: $\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$

2. Calculate $p^k$: $p^k = 0.13^3 = 0.13 \times 0.13 \times 0.13 = 0.002197$

3. Calculate $(1 - p)^{n-k}$: $(1 - p)^{n-k} = 0.87^2 = 0.87 \times 0.87 = 0.7569$

4. Multiply all parts together: $P(X = 3) = 10 \times 0.002197 \times 0.7569 = 0.01663$

Final Answer:

The probability of having exactly 3 left-handed people in the group is approximately 0.01663, or 1.663%.

Would you like more details or have any questions?

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Here are 5 related questions:

1. What is the probability of having at least 3 left-handed people in the group?
2. What is the probability of having no left-handed people in the group?
3. How does the probability change if 10 people are selected instead of 5?
4. How would you calculate the expected number of left-handed people in this group?
5. What is the probability of having exactly 2 left-handed people?

Tip: When dealing with binomial problems, always make sure you identify the number of trials, success probability, and how many successes you're looking for.