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

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

please correct me

The Gateway to Computer Science Excellence

+7 votes

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

- $\frac{1}{2}$
- $\frac{1}{7}$
- $\frac{1}{8}$
- $\frac{3}{8}$

+10 votes

Best answer

+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

please correct me

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

please correct me

+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

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

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

+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

= 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

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

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 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}$

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