SOLUTION: I have an issue with this problem for my probablity and statistics class. I hope you can help. You are dealt two cards successively from a deck of 52 playing cards. Find the p

Algebra ->  Probability-and-statistics -> SOLUTION: I have an issue with this problem for my probablity and statistics class. I hope you can help. You are dealt two cards successively from a deck of 52 playing cards. Find the p      Log On


   

Question 349042: I have an issue with this problem for my probablity and statistics class. I hope you can help.
You are dealt two cards successively from a deck of 52 playing cards. Find the probability that both cards are black in a simplified fraction.
How is the problem figured out - the answer is 25/102
Please help : )
Found 2 solutions by josmiceli, MathTherapy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
I found a way to get that answer. I'm not
positive it's right.
B = black
R = red
The cards drawn can be
BB
BR
RB
RR
The different ways to get BB or RR are
%2826%2F52%29%2A%2825%2F52%29
The different ways to get BR or RB are
%2826%2F52%29%2A%2826%2F52%29
My fraction should be BB / (BB + BR + RB + RR) =
%28%2826%2A25%29%2F52%5E2%29+%2F+%282%2A%28%2826%2A25%29%2F52%5E2+%2B+26%5E2%2F52%5E2%29%29 =
%2826%2A25%29+%2F+%282%2A%2826%2A25+%2B+26%2A26%29%29 =
25%2F%282%2A%2825+%2B+26%29%29 =
25+%2F+%282%2A51%29
25%2F102


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
I have an issue with this problem for my probablity and statistics class. I hope you can help.
You are dealt two cards successively from a deck of 52 playing cards. Find the probability that both cards are black in a simplified fraction.
How is the problem figured out - the answer is 25/102
Please help : )

There are 26 black cards in a 52-deck pack

Probability of getting a black card on 1st draw = 26%2F52

Probability of getting a black card on 2nd draw, without replacing 1st card, which could have been black = = 25%2F51

Probability of getting 2 blacks, without replacement = %2826%2F52%29+%2A+%2825%2F51%29 = highlight_green%2825%2F102%29
OR

P(black & black) = P(B) * P(B) = %2826%2F52%29+%2A+%2825%2F51%29 = highlight_green%2825%2F102%29