SOLUTION: The black face cards and the red cards numbered 3-7 are put into a bag. 4 cards are drawn at random without replacement. Find the probability that… 2 cards are black and 2 c

Algebra ->  Probability-and-statistics -> SOLUTION: The black face cards and the red cards numbered 3-7 are put into a bag. 4 cards are drawn at random without replacement. Find the probability that… 2 cards are black and 2 c      Log On


   

Question 1170141: The black face cards and the red cards numbered 3-7 are put into a bag. 4 cards are drawn at random without replacement. Find the probability that…
2 cards are black and 2 cards are red.
All cards are red.
At least 1 of the cards is black.
Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
An easier way to say this is that there are 6 black cards and 10 red cards.

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a) Probability that 2 cards are black and 2 cards are red

%286C2+%2A+10C2%29%2F%2816C4%29 = = %2815+%2A+45%29%2F1820 = 675%2F1820 = 135%2F364 = 0.3709

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b) Probability that all the cards are red

%2810C4%29%2F%2816C4%29 = %2810%21%2F%284%21%2A6%21%29%29%2F%2816%21%2F%284%21%2A12%21%29%29 = 210%2F1820 = 3%2F26 = 0.1154

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c) Probability that at least 1 of the cards is black

For this problem, you want to calculate the odds that NONE of the cards are black, then subtract this result from 1.

"None of the cards are black" is exactly the same thing as "all of the cards are red". In question 'b' above, we've already calculated the probability that all of the cards are red. So, we simply subtract this result from 1 to find out the probability that "at least 1 of the cards is black".

1+-+0.1154 = 0.8846