Determine whether events A and B are independent or dependent, then calculate P(A and B).
a. A: When a baby is born, it is a girl. 8: When a six-sided die is rolled, the result is a six.
b. A: A queen is drawn from a standard deck of cards. B: A second queen is drawn without replacement
The correct answer and explanation is :
a. Events A and B
- A: When a baby is born, it is a girl.
- B: When a six-sided die is rolled, the result is a six.
Independence of Events A and B:
Two events are independent if the occurrence of one event does not affect the occurrence of the other. In this case, the events A and B are independent because the outcome of one does not influence the other. The birth of a baby being a girl is a biological event and does not affect the result of a die roll, which is a random event.
To calculate P(A and B), which is the probability that both events occur simultaneously, we multiply the probabilities of each event:
- P(A): The probability that a baby is a girl is approximately 0.5 (since there is roughly an equal chance for a baby to be born a girl or a boy).
- P(B): The probability of rolling a six on a six-sided die is 1/6.
Thus, P(A and B)=P(A)×P(B)=0.5×16=112.P(A \text{ and } B) = P(A) \times P(B) = 0.5 \times \frac{1}{6} = \frac{1}{12}.
b. Events A and B
- A: A queen is drawn from a standard deck of cards.
- B: A second queen is drawn without replacement.
Independence of Events A and B:
In this case, events A and B are dependent because the outcome of the second draw (event B) is affected by the outcome of the first draw (event A). When a queen is drawn from the deck (event A), there are fewer queens remaining in the deck, so the probability of drawing a second queen changes.
To calculate P(A and B), we need to find the probabilities of both events occurring:
- P(A): The probability of drawing a queen from a standard deck (which has 52 cards and 4 queens) is: P(A)=452=113.P(A) = \frac{4}{52} = \frac{1}{13}.
- P(B | A): The probability of drawing a second queen, given that the first queen has already been drawn (without replacement), is: P(B∣A)=351 (since one queen is already removed, leaving 3 queens out of 51 remaining cards).P(B | A) = \frac{3}{51} \text{ (since one queen is already removed, leaving 3 queens out of 51 remaining cards)}.
Thus, P(A and B)=P(A)×P(B∣A)=113×351=3663=1221.P(A \text{ and } B) = P(A) \times P(B | A) = \frac{1}{13} \times \frac{3}{51} = \frac{3}{663} = \frac{1}{221}.
Conclusion:
- For part a, the events are independent, and the probability P(A and B)=112P(A \text{ and } B) = \frac{1}{12}.
- For part b, the events are dependent, and the probability P(A and B)=1221P(A \text{ and } B) = \frac{1}{221}.