Question 1170141
An easier way to say this is that there are 6 black cards and 10 red cards.
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<i>a) Probability that 2 cards are black and 2 cards are red</i>
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{{{(6C2 * 10C2)/(16C4)}}} = {{{(6!/(2!*4!))*(10!/(2!*8!))/(16!/(4!*12!))}}} = {{{(15 * 45)/1820}}} = {{{675/1820}}} = {{{135/364}}} = 0.3709
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<i>b) Probability that all the cards are red</i>
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{{{(10C4)/(16C4)}}} = {{{(10!/(4!*6!))/(16!/(4!*12!))}}} = {{{210/1820}}} = {{{3/26}}} = 0.1154
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<i>c) Probability that at least 1 of the cards is black
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For this problem, you want to calculate the odds that NONE of the cards are  black, then subtract this result from 1.
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"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".
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{{{1 - 0.1154}}} = 0.8846