Question-10
Students from \(3\) colleges \(X\), \(Y\), and \(Z\) have participated in a competition, where \(30\%\) of the participants are from college \(X\), \(20\%\) from college \(Y\), and \(50\%\) from college \(Z\). It is known that \(50\%\), \(40\%\) and \(60\%\) of the participants from colleges \(X\), \(Y\), and \(Z\) respectively are girls. If a girl is randomly selected, what is the probability that she is from college \(X?\) (Enter the answer correct upto three decimal places.)
Define an event as \(G\) which represents the “event that a randomly selected participant is a girl” and calculate \(P(G)\) by using law of total probability and then apply bayes theorem.
0.283
Let \(X\), \(Y\), and \(Z\) represent the events that a randomly selected participant is from college \(X\), \(Y\), and \(Z\) respectively.
Let \(G\) represent the event that a randomly selected participant is a girl.
Given, \(P(X) = 0.3, P(Y) = 0.2, P(Z) = 0.5 \hspace{2mm} \text{and,}\) \(P(G \hspace{1mm} | \hspace{1mm} X) = 0.5, P(G \hspace{1mm} | \hspace{1mm} Y) = 0.4, P(G \hspace{1mm} | \hspace{1mm} Z) = 0.6\)
Using Bayes theorem,
\(P(X \hspace{1mm} | \hspace{1mm} G) = \dfrac{P(X) P(G \hspace{1mm} | \hspace{1mm} X) }{P(G)}\)
\(P(X \hspace{1mm} | \hspace{1mm} G) = \dfrac{P(X) P(G \hspace{1mm} | \hspace{1mm} X) }{P(X) P(G \hspace{1mm} | \hspace{1mm} X) + P(Y) P(G \hspace{1mm} | \hspace{1mm} Y)+P(Z) P(G \hspace{1mm} | \hspace{1mm} Z)}\)
\(P(X \hspace{1mm} | \hspace{1mm} G) =\dfrac{(0.3 \times 0.5)}{(0.3 \times 0.5) + (0.2 \times 0.4) + (0.5 \times 0.6)} =\dfrac{0.15}{0.53} =0.283\)