Discussion
Three software engineers A, B, C are being considered for selection as project leader. The probabilities of their selection are in the ratio 1 : 3 : 6. If A is selected, the probability that she will use R software is 0.8. The corresponding probabilities for B and C are respectively 0.5 and 0.3. The following are two statements :
(I) The probability that R software will be used for the new project is 0.41.
(II) If R software is used, the probability that either A or B is the project leader is 23/41.
Which of the following is true ?
Answer: A
Both (I) and (II) are true.
The selection probabilities are \(\frac{1}{10}, \frac{3}{10}, \frac{6}{10}\). By the law of total probability,
\[ \begin{aligned} P(R) &= 0.1(0.8) + 0.3(0.5) + 0.6(0.3) \\ &= 0.08 + 0.15 + 0.18 = 0.41 \end{aligned} \]
which is (I). By Bayes' theorem,
\[ \begin{aligned} P(A \text{ or } B \mid R) &= \frac{0.08 + 0.15}{0.41} \\ &= \frac{23}{41} \end{aligned} \]
which is (II). The remaining \(\frac{18}{41}\) is the chance that C is the leader given that R is used.
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