This problem involves finding probabilities for a binomial distribution, as we are dealing with two outcomes (left-handed or not) with a fixed probability of success for each trial.

Given:

• Probability of a person being left-handed, $p = 0.13$
• Probability of a person being right-handed, $q = 1 - p = 0.87$
• Sample size: 6 customers.

Let's break it down step by step.

Part a) The first lefty is the fifth person chosen.

This situation can be modeled as a geometric distribution, where we are waiting for the first "success" (lefty) on the 5th trial.

The probability of the first lefty being the 5th person chosen is: $P(\text{first lefty on the 5th person}) = (0.87)^4 \times 0.13$

Since this answer is already provided as 0.0745, let's move to part b.

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Part b) There are some lefties among the 6 people.

This requires finding the probability of at least one left-handed person among the 6 customers. We can solve this using the complement rule: $P(\text{at least 1 lefty}) = 1 - P(\text{no lefties})$

The probability of no lefties among the 6 customers is: $P(\text{no lefties}) = (0.87)^6$

So, the probability of at least one lefty is: $P(\text{at least 1 lefty}) = 1 - (0.87)^6$

Let me calculate that value.The probability that there are some left-handed individuals among the 6 people is approximately 0.5664 (rounded to four decimal places).

Would you like more details or have any questions?

Here are five related questions to deepen your understanding:

1. How would you calculate the probability of exactly two lefties among the 6 people?
2. What is the expected number of lefties in a sample of 10 people?
3. How would the probability change if the sample size increased to 10 customers?
4. Can you calculate the probability that the first lefty is the third person chosen?
5. What is the probability that all 6 people are left-handed?

Tip: For problems involving "at least one," using the complement rule (1 minus the probability of none) is often easier than summing probabilities for each individual outcome.