To solve this problem, we need to use concepts from probability theory. Specifically, we'll use the binomial distribution and geometric distribution to calculate the probabilities for each part.

Part (e): Probability that there are at least 3 lefties in the group

We need to calculate the probability of having 3 or more left-handed people in a group of 6. This can be done using the binomial probability formula:

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

where:

• $n = 6$ (number of trials, i.e., number of people selected),
• $p = 0.13$ (probability of success, i.e., selecting a left-handed person),
• $k$ is the number of successes.

To find the probability of "at least 3 lefties," we sum up the probabilities for $k = 3, 4, 5,$ and $6$.

$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

Let's calculate this step-by-step:

1. Calculate $P(X = 3)$: $P(X = 3) = \binom{6}{3} (0.13)^3 (0.87)^3$

2. Calculate $P(X = 4)$: $P(X = 4) = \binom{6}{4} (0.13)^4 (0.87)^2$

3. Calculate $P(X = 5)$: $P(X = 5) = \binom{6}{5} (0.13)^5 (0.87)^1$

4. Calculate $P(X = 6)$: $P(X = 6) = \binom{6}{6} (0.13)^6 (0.87)^0$

5. Sum the probabilities: $P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

Let's perform these calculations.The probability that there are at least 3 lefties in the group is approximately $0.0324$ (rounded to four decimal places).

Do you want further details or have any questions about this calculation?

Related Questions:

1. What is the probability of having exactly 2 left-handed people in the group?
2. How would the probability change if the sample size were increased to 10 people?
3. What is the probability of having no left-handed people in the group?
4. How can we use the geometric distribution to calculate the probability that the first lefty is found on the fifth attempt?
5. What is the expected number of left-handed people in a sample of 6?

Tip: When dealing with "at least" probabilities, it is often easier to calculate the complementary "less than" probabilities and subtract from 1.