Three cards are drawn at random without replacement from a standard deck of 52 playing cards. What is the probability that the second and third cards are black?
There were 52 cards. Half were black
so
26/52 = .5
then we had 51 cards, 25 black
25/51
then we had 50 cards, 24 black
24/50
so
.5 * 25/51 * 24/50
Did not say that the first card could have been black also, so ..
26 Black , 26 Red
case 1, the first card was also black, that is, BBB
prob of that = (26/52)(25/51)(24/50)
case 2, the first card was not black, but the other two were black, that is, RBB
prob of that = (26/52)(26/51)(25/50)
prob of your event = (26/52)(25/51)(24/50) + (26/52)(26/51)(25/50)
= (26/52)((25/51)(24/50) + (26/51)(25/50)
= (26/52)(25*24 + 26*25)/(50(51))
= ....
Oh, thanks, did not notice that.
To find the probability that the second and third cards drawn are black, we need to consider the number of possible favorable outcomes and the number of possible total outcomes.
First, let's consider the number of possible favorable outcomes. There are 26 black cards in a standard deck of 52 playing cards. For the second card, after drawing one card, there are 51 cards remaining, and out of those, 25 are black. Therefore, the probability of drawing a black card as the second card is 25/51.
Now, for the third card, after drawing two cards, there are 50 cards remaining, and out of those, 24 are black. Therefore, the probability of drawing a black card as the third card is 24/50.
To find the probability of both events occurring, we multiply the probabilities together: (25/51) * (24/50) = 24/102.
So, the probability that the second and third cards drawn are black is 24/102, which can be simplified to 12/51 or approximately 0.235.