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I solved the problem. I am getting the final answer as 8/15, however, the book says the answer is 7/15.Seems like they haven't subtracted 7/15 from 1 Please help.

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    $\begingroup$ Can you show your calculations? $\endgroup$ – Bram28 Oct 17 '17 at 19:28
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Let $B$ be the event that a person is a boy. Let $M$ be the event that a person score more than $40$ marks. We are told $Pr(B)=0.6, Pr(M)=0.6, Pr(M\mid B^c)=0.8$. We are tasked with calculating $Pr(M^c\mid B)$.

Continuing to build a list of possibly useful numbers: $Pr(M\cap B^c)=Pr(B^c)Pr(M\mid B^c)=0.4\cdot 0.8=0.32$

$Pr(M\cap B)=Pr(M)-Pr(M\cap B^c)=0.6-0.32=0.28$

$Pr(M^c\cap B)=Pr(B)-Pr(M\cap B)=0.6-0.28=0.32$

$Pr(M^c\mid B)=\frac{Pr(M^c\cap B)}{Pr(B)}=\frac{0.32}{0.6}=\frac{8}{15}$

My answer agrees with yours.

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We have $60\%$ boys and $40\%$ girls.

We know $20\%$ of the girls have less equal $40$ points, these are $8\%$ of the students.

We know $40\%$ of the students have less equal $40$ points.

So the boys which have $40$ or less points are $32\%$ of the students, these are $32/60 =8/15$ of the boys.

Hm, that information about the range $0$ to $150$ marks seems not to matter. No idea how to reduce to $7/15$.

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Let there be 100 students in a class.

60 are boys, 40 girls.

32 girls scored more than 40 points.

60 boys and girls scored more than 40.

$\rightarrow$: 28 boys scored more than 40.

$\rightarrow $: 32 boys scored 40 or less.

Fraction : 32/60= 8/15.

Answer: 8/15 of the boys scored 40 marks or less.

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