Question

In a class of 75 students, 15 are above average, 45 are average and the rest below average achievers. The probability that an above average achieving student fails is 0.005, that an average achieving student fails is 0.05 and the probability of a below average achieving student failing is 0.15. If a student is know to have passed, what is the probability that he is a below average achiever?

Answer 1

Let, $E_1$ : Event that student is above average
$E_2$ : Event that student is average
$E_3$ : Event that student is below average.
A : Event that student is known to have passed
$P\left(E_1\right)=\frac{1}{5},P\left(E_2\right)=\frac{3}{5},P\left(E_3\right)=\frac{1}{5}$
$P\left(A/E_1\right)=1-0.005=0.995$
$P\left(A/E_2\right)=1-0.05=0.95$
$P\left(A/E_3\right)=1-0.15=0.85$
Using Baye's theorem,
$P\left(E_3/A\right)=\frac{P\left(E_3\right).P\left(A/E_3\right)}{P\left(E_1\right).P\left(A/E_1\right)+P\left(E_2\right).P\left(A/E_2\right)+P\left(E_3\right).P\left(A/E_3\right)}$

$=\frac{\frac{1}{5}\times \frac{85}{100}}{\frac{1}{5}\times \frac{995}{1000}+\frac{3}{5}\times \frac{95}{100}+\frac{1}{5}\times \frac{85}{100}}=\frac{\frac{17}{100}}{\frac{199}{1000}+\frac{57}{100}+\frac{17}{100}}$

$=\frac{\frac{17}{100}}{\frac{199+570+170}{1000}}$

$=\frac{\frac{17}{100}}{\frac{939}{1000}}=\frac{17\times 100}{100\times 939}=\frac{170}{939}=0.1810$