Asked by Hannah
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?
Answers
Answered by
Damon
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
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
Answered by
Reiny
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))
= ....
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))
= ....
Answered by
Damon
Oh, thanks, did not notice that.