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question 34 (1 point) saved exhibit 7 - 8 a standard 52 - card deck is …

Question

question 34 (1 point) saved exhibit 7 - 8 a standard 52 - card deck is shuffled and 2 cards are picked from the top of the deck. refer to exhibit 7 - 8. the probability that both cards are hearts is 6.3% 5.9% 6.4% 25.0% question 35 (1 point) saved exhibit 7 - 8 a standard 52 - card deck is shuffled and 2 cards are picked from the top of the deck. refer to exhibit 7 - 8. the probability that the first card is a face card (jack, queen, king) and the second card is not a face card is

Explanation:

Step1: Calculate probability of first - card being a face - card

There are 12 face - cards in a 52 - card deck. So the probability that the first card is a face - card, $P(F_1)=\frac{12}{52}=\frac{3}{13}$.

Step2: Calculate probability of second - card not being a face - card given first is a face - card

After drawing a face - card first, there are 51 cards left, and 40 non - face cards. So the probability that the second card is not a face - card given the first is a face - card, $P(\overline{F_2}|F_1)=\frac{40}{51}$.

Step3: Use the multiplication rule for conditional probability

The probability that the first card is a face - card and the second is not a face - card is $P = P(F_1)\times P(\overline{F_2}|F_1)=\frac{3}{13}\times\frac{40}{51}=\frac{120}{663}\approx 0.181 = 18.1\%$.

However, it seems the main question here is about the probability that both cards are hearts.

Step1: Calculate probability of first card being a heart

There are 13 hearts in a 52 - card deck. So the probability that the first card is a heart, $P(H_1)=\frac{13}{52}=\frac{1}{4}$.

Step2: Calculate probability of second card being a heart given first is a heart

After drawing a heart first, there are 51 cards left and 12 hearts left. So the probability that the second card is a heart given the first is a heart, $P(H_2|H_1)=\frac{12}{51}$.

Step3: Use the multiplication rule for conditional probability

The probability that both cards are hearts is $P = P(H_1)\times P(H_2|H_1)=\frac{13}{52}\times\frac{12}{51}=\frac{156}{2652}=\frac{1}{17}\approx 0.059 = 5.9\%$

Answer:

5.9%