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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. $\frac{3}{8}$
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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
+1
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

+1
$(M,F,F) \rightarrow$ 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
0
Hope you understand
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yes. thanks
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nice @leensharma
+1
why order is important in this question??? here it is not mentioned that which one numbered should be male or female????
0
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?
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
0
Awesome. More intuitive.
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
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

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

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

=> $\frac{1}{7}$
–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
+2
given that At least one is male.
+3
I think Ans should be - B (1/7)
+3
@srestha F,F,F combination is not possible in sample space, as given condition is that at least one male.
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yes :)