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8 votes
8 votes

A man has three cats. At least one is male. What is the probability that all three are male?

  1. $\frac{1}{2}$
  2. $\frac{1}{7}$
  3. $\frac{1}{8}$
  4. $\frac{3}{8}$

5 Answers

Best answer
10 votes
10 votes

Answer is (B)

Given that $A$ man has three cats and  At least one is male.

Possible combination for at least one cat is male=$(M,F,F),(F,M,F),(F,F,M),(M,F,M),(M,M,F),(F,M,M),(M,M,M)$

Probability that all three are male=$\frac{1}{7}$

 

edited by
15 votes
15 votes
P(all 3 males / atleast one male) = P(all 3 males AND atleast one male) / P(atleast one male)

                                                    = P(all 3 males) / P(atleast one male)

                                                    = P( (all 3 males) / [1-P(no males] )

                                                    = (1/8) / (1-1/8)

                                                   = (1/8) / (7/8)

                                                   = 1/7
0 votes
0 votes
Option B is the correct

It is a que of conditional probability

so probability that atleast one is male A = c(3,1)/8

probability that atleast two are male B=c(3,2)/8

probability that atleast three male C= c(3,3)/8

p(atleast one is male)=A+B+C/8 =7/8

All 3 are male

p(All 3 are male)=1/8

so total prob =p(All 3 are male)/p(atleast one is male) = 1/7
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Let A be the event that at least one cat is Male. Let B be the event that all the three cats are Male.

By conditional probability:

$P(B|A) = \frac{P(B\cap A)}{P(A)}$

 

=> If you think about it, $P(B\cap A)=P(B)$ because $A \subset B$

So, our equation is reduced to

$P(B|A) = \frac{P(B)}{P(A)}$

=> $ \frac{1/8}{7/8}$

=> $ \frac{1}{7}$
Answer:

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