Asked by ana
A litter of eight puppies consists of 5 males and 3 females. Two puppies are chosen at random, without replacement. What is the probability that both puppies will be female?
Answers
Answered by
mathhelper
<b>prob(your stated event) = C(3,2)/C(8,2) = 3/28</b>
btw, breakdown of probabilities:
both females = 3/28 (see above)
both males = C(5,2)/C(8,2) = 10/28
one male, one female = C(5,1)*C(3,1)/C(8,2 = 15/28
note 3/28 + 10/28 + 15/28 = 28/28 = 1
btw, breakdown of probabilities:
both females = 3/28 (see above)
both males = C(5,2)/C(8,2) = 10/28
one male, one female = C(5,1)*C(3,1)/C(8,2 = 15/28
note 3/28 + 10/28 + 15/28 = 28/28 = 1
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