Question
Michael draws a card from a standard deck of 52 cards (and puts it back) 20 times.
How many times can Michael expect to draw a diamond?
How many times can Michael expect to draw a diamond?
Answers
GPT-4o mini
A standard deck of cards contains 52 cards, with 13 cards in each suit (hearts, diamonds, clubs, spades). Therefore, the probability of drawing a diamond in a single draw is:
\[
P(\text{diamond}) = \frac{\text{number of diamonds}}{\text{total number of cards}} = \frac{13}{52} = \frac{1}{4}
\]
Michael draws a card 20 times, and we want to find the expected number of times he draws a diamond. The expected value \(E\) for a binomial distribution can be calculated using the formula:
\[
E = n \cdot p
\]
where \(n\) is the number of trials (draws) and \(p\) is the probability of success (drawing a diamond).
In this case, we have:
- \(n = 20\) (the number of draws)
- \(p = \frac{1}{4}\) (the probability of drawing a diamond)
Now we can calculate the expected number of diamonds drawn:
\[
E = 20 \cdot \frac{1}{4} = 5
\]
Therefore, Michael can expect to draw a diamond **5 times**.
\[
P(\text{diamond}) = \frac{\text{number of diamonds}}{\text{total number of cards}} = \frac{13}{52} = \frac{1}{4}
\]
Michael draws a card 20 times, and we want to find the expected number of times he draws a diamond. The expected value \(E\) for a binomial distribution can be calculated using the formula:
\[
E = n \cdot p
\]
where \(n\) is the number of trials (draws) and \(p\) is the probability of success (drawing a diamond).
In this case, we have:
- \(n = 20\) (the number of draws)
- \(p = \frac{1}{4}\) (the probability of drawing a diamond)
Now we can calculate the expected number of diamonds drawn:
\[
E = 20 \cdot \frac{1}{4} = 5
\]
Therefore, Michael can expect to draw a diamond **5 times**.