Question

A software company employs 3 programmers A, B and C. For the duration of a particular project, A, B and C worked 30%, 50% and 20% respectively. The probability of python programming language being used by A, B and C is 5%, 10% and 15% respectively. Find the probability that a particular piece of code chosen randomly is written by A given that it is in python language.

• A. 0.157
• B. 0.239
• C. 0.175
• D. 0.186

Answer 1

The correct option is A 0.157

$\mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{the}\mathrm{following}\mathrm{events}:D:‘\mathrm{the}\mathrm{code}\mathrm{is}\mathrm{in}\mathrm{python}\mathrm{programming}\mathrm{language}’.E_1:‘A\mathrm{worked}\mathrm{on}\mathrm{the}\mathrm{project}’.E_2:‘B\mathrm{worked}\mathrm{on}\mathrm{the}\mathrm{project}’.E_3:‘C\mathrm{worked}\mathrm{on}\mathrm{the}\mathrm{project}’.$

$\mathrm{Then}P\left(E_1\right)=0.3,P\left(E_2\right)=0.5,P\left(E_3\right)=0.2\mathrm{and}P\left(D|E_1\right)=0.05,P\left(D|E_2\right)=0.1,P\left(D|E_3\right)=0.15$

$\mathrm{We}\mathrm{can}\mathrm{infer}\mathrm{that}{E}_1,E_2\mathrm{and}{E}_3\mathrm{are}\mathrm{mutually}\mathrm{exclusive}\mathrm{and}\mathrm{exhaustive}\mathrm{events}\mathrm{and}\mathrm{hence},\mathrm{represent}\mathrm{partition}\mathrm{of}\mathrm{sample}\mathrm{space}.\mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}P\left(E_1|D\right).\mathrm{By}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{have}P\left(E_1|D\right)=\frac{P\left(D|E_1\right)P\left(E_1\right)}{P\left(D|E_1\right)P\left(E_1\right)+P\left(D|E_2\right)P\left(E_2\right)+P\left(D|E_3\right)P\left(E_3\right)}\Rightarrow P\left(E_1|D\right)=\begin{array}{ccc}& & \frac{0.05\times 0.3}{0.05\times 0.3+0.1\times 0.5+0.15\times 0.2}\end{array}\Rightarrow P\left(E_1|D\right)=\begin{array}{ccc}& & \frac{0.015}{0.015+0.05+0.03}\end{array}\Rightarrow P\left(E_1|D\right)=\frac{15}{95}=\begin{array}{ccc}& & 0\end{array}\begin{array}{ccc}& & .\end{array}\begin{array}{ccc}& & 157\end{array}\mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{probability}\mathrm{is}0.157.$