Question
Margarita draws two cards from a standard deck of cards. Event A
is to draw any card from the suit of hearts without replacing it and event B
is to draw any red card. What is P(A∩B)
? Note that there are 26 red cards in a standard deck and that a suit includes 13 cards. Round your answer to the nearest tenths place.(1 point)
is to draw any card from the suit of hearts without replacing it and event B
is to draw any red card. What is P(A∩B)
? Note that there are 26 red cards in a standard deck and that a suit includes 13 cards. Round your answer to the nearest tenths place.(1 point)
Answers
GPT-4o mini
To find \( P(A \cap B) \), we need to examine what events A and B entail:
- Event A: Drawing a heart card. There are 13 heart cards in a standard deck of 52 cards.
- Event B: Drawing a red card. There are 26 red cards total (13 hearts and 13 diamonds).
The intersection \( A \cap B \) represents drawing a card that is both a heart and a red card. Since all heart cards are red, drawing a heart card satisfies the condition for drawing a red card. Thus, the scenario of drawing a heart card is identical to drawing a card from event B, given it is a red card.
1. **Computing the probabilities**:
- For the first card drawn (let's call it card 1), the probability of drawing a heart (event A) is \( P(A) = \frac{13}{52} = \frac{1}{4} \).
- If card 1 is a heart, we do not replace it, leaving us with 51 cards in total.
2. **Finding \( P(B|A) \)**:
- If you have drawn a heart first (card 1), there are now 25 red cards left in the remaining 51 cards (12 hearts left and 13 diamonds = 25).
- So, \( P(B|A) = \frac{25}{51} \).
3. **Finding \( P(A \cap B) = P(A) \times P(B|A) \)**:
\[
P(A \cap B) = P(A) \times P(B|A) = \frac{13}{52} \times \frac{25}{51}
\]
\[
P(A \cap B) = \frac{1}{4} \times \frac{25}{51} = \frac{25}{204}
\]
4. **Calculating the final probability**:
To convert \( \frac{25}{204} \) to decimal form:
\[
\frac{25}{204} \approx 0.1225
\]
Rounding to the nearest tenths place gives:
\[
P(A \cap B) \approx 0.1
\]
Thus, the answer is \( \boxed{0.1} \).
- Event A: Drawing a heart card. There are 13 heart cards in a standard deck of 52 cards.
- Event B: Drawing a red card. There are 26 red cards total (13 hearts and 13 diamonds).
The intersection \( A \cap B \) represents drawing a card that is both a heart and a red card. Since all heart cards are red, drawing a heart card satisfies the condition for drawing a red card. Thus, the scenario of drawing a heart card is identical to drawing a card from event B, given it is a red card.
1. **Computing the probabilities**:
- For the first card drawn (let's call it card 1), the probability of drawing a heart (event A) is \( P(A) = \frac{13}{52} = \frac{1}{4} \).
- If card 1 is a heart, we do not replace it, leaving us with 51 cards in total.
2. **Finding \( P(B|A) \)**:
- If you have drawn a heart first (card 1), there are now 25 red cards left in the remaining 51 cards (12 hearts left and 13 diamonds = 25).
- So, \( P(B|A) = \frac{25}{51} \).
3. **Finding \( P(A \cap B) = P(A) \times P(B|A) \)**:
\[
P(A \cap B) = P(A) \times P(B|A) = \frac{13}{52} \times \frac{25}{51}
\]
\[
P(A \cap B) = \frac{1}{4} \times \frac{25}{51} = \frac{25}{204}
\]
4. **Calculating the final probability**:
To convert \( \frac{25}{204} \) to decimal form:
\[
\frac{25}{204} \approx 0.1225
\]
Rounding to the nearest tenths place gives:
\[
P(A \cap B) \approx 0.1
\]
Thus, the answer is \( \boxed{0.1} \).