Question

A deck of 33 cards contains 8 red cards, 10 white cards, and 15 blue cards. For this problem we will select a sample of 6 cards at random with replacement. That is we will select the cards one at a time returning the selected card to the deck before selecting the next card. Hint: When you are counting favorable outcomes for this problem you need to account for the order in which the cards are selected. a) Find the probability that among the 6 cards selected there are exactly 2 red cards. b) Find the probability that among the 6 cards selected there are exactly 4 red cards. c) Find the probability that among the 6 cards selected there are 2 red cards and 4 white cards. d) Find the probability that among the 6 cards selected there are 1 red card, 2 white cards, and 3 blue cards.

Answer 1

a) To find the probability that among the 6 cards selected there are exactly 2 red cards, we need to consider the number of ways we can choose 2 red cards (out of 8) and 4 non-red cards (out of 25) from the deck with replacement.

The probability of selecting a red card is 8/33, and the probability of selecting a non-red card is (25/33). Since we are selecting cards with replacement, the probabilities remain the same for each draw.

Using the concept of combinations, the probability can be calculated as follows:

P(2 red cards) = C(6, 2) * (8/33)^2 * (25/33)^4

where C(6, 2) represents the number of combinations of choosing 2 items from 6.

Plugging in the values:

P(2 red cards) = (6! / (2! * (6-2)!)) * (8/33)^2 * (25/33)^4

= 15 * (8/33)^2 * (25/33)^4

≈ 0.2098 (rounded to four decimal places)

Therefore, the probability of selecting exactly 2 red cards among the 6 cards is approximately 0.2098.

b) To find the probability that among the 6 cards selected there are exactly 4 red cards, we follow a similar approach.

P(4 red cards) = C(6, 4) * (8/33)^4 * (25/33)^2

Plugging in the values:

P(4 red cards) = (6! / (4! * (6-4)!)) * (8/33)^4 * (25/33)^2

= 15 * (8/33)^4 * (25/33)^2

≈ 0.0958 (rounded to four decimal places)

Therefore, the probability of selecting exactly 4 red cards among the 6 cards is approximately 0.0958.

c) To find the probability that among the 6 cards selected there are 2 red cards and 4 white cards, we once again use the concept of combinations along with the probabilities.

P(2 red cards and 4 white cards) = C(6, 2) * (8/33)^2 * (10/33)^4

Plugging in the values:

P(2 red cards and 4 white cards) = (6! / (2! * (6-2)!)) * (8/33)^2 * (10/33)^4

= 15 * (8/33)^2 * (10/33)^4

≈ 0.1790 (rounded to four decimal places)

Therefore, the probability of selecting exactly 2 red cards and 4 white cards among the 6 cards is approximately 0.1790.

d) To find the probability that among the 6 cards selected there are 1 red card, 2 white cards, and 3 blue cards, we follow a similar approach.

P(1 red card, 2 white cards, and 3 blue cards) = C(6, 1) * C(5, 2) * (8/33) * (10/33)^2 * (15/33)^3

Plugging in the values:

P(1 red card, 2 white cards, and 3 blue cards) = (6! / (1! * (6-1)!)) * (5! / (2! * (5-2)!)) * (8/33) * (10/33)^2 * (15/33)^3

= 6 * 10 * (8/33) * (10/33)^2 * (15/33)^3

≈ 0.2089 (rounded to four decimal places)

Therefore, the probability of selecting exactly 1 red card, 2 white cards, and 3 blue cards among the 6 cards is approximately 0.2089.