A manufacturer of game controllers is concerned that its controller may be difficult for​ left-handed users. They set out to find lefties to test. About 13​% of the population is​ left-handed. If they select a sample of 6 customers at random in their​ stores, what is the probability of each of the outcomes described in parts a through f​ below?
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Part 1
​a) The first lefty is the fifth person chosen.
The probability is 
  
0.0745.
​(Round to four decimal places as​ needed.)
Part 2
​b) There are some lefties among the 6 people.
The probability is 
  
0.5664.
​(Round to four decimal places as​ needed.)
Part 3
​c) The first lefty is the second or third person.
The probability is 
  
0.2115.
​(Round to four decimal places as​ needed.)
Part 4
​d) There are exactly 3 lefties in the group.
The probability is 
  
0.0289.
​(Round to four decimal places as​ needed.)
Part 5
​e) There are at least 3 lefties in the group.
The probability is 
  
enter your response here.
​(Round to four decimal places as​ needed.)

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.

A manufacturer of game controllers is concerned that its controller may be difficult for left-handed users. They set out to find lefties to test. About 13% of the population is left-handed. If they select a sample of 6 customers at random in their stores, what is the probability of each of the outcomes described in parts a through f below?

Explore the probability of selecting left-handed users from a random sample of 6 people, using binomial and geometric distributions. This problem involves calculating probabilities for various outcomes, such as the probability that at least 3 people are left-handed or that the first lefty appears on the fifth selection. Suitable for college-level probability students or advanced high school students.

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