Question
Xander spends most of his time with his 10 closest friends. He has known 4 of his 10 friends since kindergarten. If he is going to see a movie tonight with 3 of his 10 closest friends, what is the probability that the first 2 of the friends to show up to the movie are friends he has known since kindergarten but the third is not?
StartFraction 1 over 30 EndFraction
StartFraction 9 over 125 EndFraction
StartFraction 12 over 125 EndFraction
StartFraction 1 over 10 EndFraction
StartFraction 1 over 30 EndFraction
StartFraction 9 over 125 EndFraction
StartFraction 12 over 125 EndFraction
StartFraction 1 over 10 EndFraction
Answers
GPT-4o mini
To determine the probability that the first 2 friends to show up are ones he has known since kindergarten and the third friend is not, we can break down the situation as follows:
1. **Identify the number of friends**:
- Total friends = 10
- Friends known since kindergarten = 4
- Friends not known since kindergarten = 10 - 4 = 6
2. **Calculate the probability for the specific sequence**: We need to find the probability that the first friend is a kindergarten friend, the second friend is also a kindergarten friend, and the third friend is not a kindergarten friend.
- **Choosing the first friend**: The probability that the first friend is a kindergarten friend is:
\[
P(\text{1st is kindergarten}) = \frac{4}{10}
\]
- **Choosing the second friend**: After the first friend who is a kindergarten friend shows up, we have 3 kindergarten friends left and a total of 9 friends to choose from. The probability that the second friend is also a kindergarten friend is:
\[
P(\text{2nd is kindergarten | 1st is kindergarten}) = \frac{3}{9} = \frac{1}{3}
\]
- **Choosing the third friend**: After the first two friends (both kindergarten friends) show up, we have 6 friends not known since kindergarten, and a total of 8 friends left. The probability that the third friend is not a kindergarten friend is:
\[
P(\text{3rd is not kindergarten | 1st and 2nd are kindergarten}) = \frac{6}{8} = \frac{3}{4}
\]
3. **Combine the probabilities**: Now we need to multiply these probabilities together to find the total probability for this specific sequence:
\[
P(\text{1st is kindergarten}) \times P(\text{2nd is kindergarten | 1st is kindergarten}) \times P(\text{3rd is not kindergarten | 1st and 2nd are kindergarten})
\]
\[
= \frac{4}{10} \times \frac{1}{3} \times \frac{3}{4}
\]
Now calculate the product:
\[
= \frac{4}{10} \times \frac{1}{3} \times \frac{3}{4} = \frac{4 \times 1 \times 3}{10 \times 3 \times 4} = \frac{12}{120} = \frac{1}{10}
\]
Thus, the probability that the first two friends to show up are friends he has known since kindergarten and the third is not is:
\[
\boxed{\frac{1}{10}}
\]
1. **Identify the number of friends**:
- Total friends = 10
- Friends known since kindergarten = 4
- Friends not known since kindergarten = 10 - 4 = 6
2. **Calculate the probability for the specific sequence**: We need to find the probability that the first friend is a kindergarten friend, the second friend is also a kindergarten friend, and the third friend is not a kindergarten friend.
- **Choosing the first friend**: The probability that the first friend is a kindergarten friend is:
\[
P(\text{1st is kindergarten}) = \frac{4}{10}
\]
- **Choosing the second friend**: After the first friend who is a kindergarten friend shows up, we have 3 kindergarten friends left and a total of 9 friends to choose from. The probability that the second friend is also a kindergarten friend is:
\[
P(\text{2nd is kindergarten | 1st is kindergarten}) = \frac{3}{9} = \frac{1}{3}
\]
- **Choosing the third friend**: After the first two friends (both kindergarten friends) show up, we have 6 friends not known since kindergarten, and a total of 8 friends left. The probability that the third friend is not a kindergarten friend is:
\[
P(\text{3rd is not kindergarten | 1st and 2nd are kindergarten}) = \frac{6}{8} = \frac{3}{4}
\]
3. **Combine the probabilities**: Now we need to multiply these probabilities together to find the total probability for this specific sequence:
\[
P(\text{1st is kindergarten}) \times P(\text{2nd is kindergarten | 1st is kindergarten}) \times P(\text{3rd is not kindergarten | 1st and 2nd are kindergarten})
\]
\[
= \frac{4}{10} \times \frac{1}{3} \times \frac{3}{4}
\]
Now calculate the product:
\[
= \frac{4}{10} \times \frac{1}{3} \times \frac{3}{4} = \frac{4 \times 1 \times 3}{10 \times 3 \times 4} = \frac{12}{120} = \frac{1}{10}
\]
Thus, the probability that the first two friends to show up are friends he has known since kindergarten and the third is not is:
\[
\boxed{\frac{1}{10}}
\]