Question 1074771: a bag contains 6 white,4 red and 4 black balls. find the probability of getting at least two red balls if 4 balls are drawn.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Probability of 0 red balls: (10/14)(9/13)(8/12)(7/11), since 10 of 14 are not red. This is 0.2098.
Probability of 1 red ball: The denominator is the same. The numerator is 4*10*9*8*4; the first 4 is the number of ways the 1 ball can appear. The 10,9,8 are the non-red balls, and the 4 is the number of red balls present. The order or the numbers in the numerator can change, but multiplication is commutative.
That probability is 0.4795
At least 2 is 1- the sum of these two probabilities or 0.3107.
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Check by doing 2, which is 6 ways 2 red balls can be chosen out of 4, and the numerator is 6*4*3*10*9=6480.
For 3, it is 4*10*4*3*2=960
for 4, it is 4*3*2*1=24
That sum is 7464/(14*13*12*11)=0.3107.
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