From a standard deck of cards, two cards are randomly drawn, one after the other without replacement. The color of each car is noted. Which is less likely to occur: two red cards or a red card and a black card? Explain.
prob of 2 red cards = (26/52)(25/51) = 25/102
prob of red and black = 2(26/52)(26/51) = 26/51
or
prob two reds = C(26,2)/C(52,2) = 25/102
prob rd and black = C(26,1) x C(26,1)/C(52,2) = 26/51
To determine which outcome is less likely to occur, we need to compare the probabilities of drawing two red cards and drawing a red card and a black card.
Let's break down the two scenarios:
1. Two red cards:
To compute the probability of drawing two red cards, we first determine the probability of drawing a red card for the first draw, which would be 26 red cards out of 52 total cards. After the first card is drawn, there are now 25 red cards remaining out of 51 total cards for the second draw. Therefore, the probability of drawing two red cards is:
(26/52) * (25/51) = 325/2652 ≈ 0.1224
2. A red card and a black card:
For this scenario, we calculate the probability of drawing a red card followed by a black card in any order. There are two possibilities to consider: drawing a red card first and then a black card, or drawing a black card first and then a red card.
The probability of drawing a red card followed by a black card would be (26/52) * (26/51), because after drawing a red card, there are 26 black cards remaining out of the remaining 51 cards. The probability of drawing a black card followed by a red card would be (26/52) * (26/51) as well, for the same reason.
Adding these two probabilities together gives us:
(26/52) * (26/51) + (26/52) * (26/51) = 2 * (26/52) * (26/51) = 676/2652 ≈ 0.2551
Comparing the two probabilities, we find that the probability of drawing two red cards is approximately 0.1224, while the probability of drawing a red card and a black card is approximately 0.2551. Therefore, it is less likely to occur that we draw two red cards compared to drawing a red card and a black card.