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+7 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. $\frac{3}{8}$
in Probability by Veteran | 579 views

5 Answers

+10 votes
Best answer

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=$\frac{1}{7}$


by Boss
edited by
Here, why we are counting two different counts for same result?

i think Eg (M,F,F) ,(F,M,F), (F,F,M) all should be same

please correct me
$(M,F,F) \rightarrow $ Here 1st cat is male, 2nd is female, 3rd is also female

$(F,M,F) \rightarrow $ Here 1st & 3rd is female & 2nd is male

$(F,F,M) \rightarrow $ Here 1st & 2nd is female & 3rd are male
Hope you understand
yes. thanks
nice @leensharma
why order is important in this question??? here it is not mentioned that which one numbered should be male or female????
How the order got importance in this question?
After all the man owns three cats and one of them is male then $(M,F,F) ,(F,M,F), (F,F,M)$  is same isn't it?
+10 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
by Loyal
Awesome. More intuitive.
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
0 votes
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}$
by Loyal
–1 vote
Probability of cat is either male or female (F,F,F),(M,F,F),(F,M,F),(F,F,M),(M,F,M),(M,M,F),(F,M,M),(M,M,M)

Probability  of all three are male is (M,M,M) =1/8

Ans C)
by Veteran
given that At least one is male.
I think Ans should be - B (1/7)
@srestha F,F,F combination is not possible in sample space, as given condition is that at least one male.
yes :)

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