Can You check pls ?
Problem: there are 110 girls and 90 boys in the class,30% girls and 40% boys go to the school by bicycle.Probability ,that a boy gets into an accident is twice as much as a girl.
(i) find a probability ,that accident kind is a girl
(ii) let a probability,that a girl gets into accident is 0.02 . Find the expected number of childen,who gets into an accident.
My idea:
(i):Let G=girls,B=boys,A-accident.x-probability girls accident than 2x -boys. than probability
$$P(G|A)=\frac{P(A|G)*P(G) }{ P(A|G)*P(G)+P(A|G^c)*P(G^c)}$$
$=0.3*x*0.45/(0.3x*0.45+0.55*0.4*2x)=0.23$
(ii)
$N=0.02*0.3+110+0.04*0.4*90=2.1$ ,about 2 children
Thank You
I don't think your solution to part (i) is correct.
We are told that $33$ girls and $36$ boys go to school by bicycle, so $$\Pr(G)=\frac{33}{69}$$ Also $\Pr(A|G)=x$ and $\Pr(A|B)=2x$. Therefore, $$\Pr(G|A)=\frac{\frac{33}{69}x}{\frac{33}{69}x+\frac{36}{69}\cdot2x}=\frac{33}{33+72}=\frac{11}{35}\approx.3142857$$