Discussion
In 1989 there were three candidates for the position of Principal, Mr. Chatterji, Mr. Ayangar and Dr. Singh, whose chances of getting the appointment are in the proportion 4 : 2 : 3 respectively. The probability that Mr. Chatterji if selected would introduce co-education in the college is 0.3. The probabilities of Mr. Ayangar and Dr. Singh doing the same are respectively 0.5 and 0.8. What is the probability that there was co-education in the college in 1990 ?
Answer: D
The probability is \(\frac{23}{45}\).
The chances 4 : 2 : 3 give selection probabilities of 4/9, 2/9 and 3/9. By the theorem of total probability:
\[ \begin{aligned} P &= \tfrac{4}{9}(0.3) + \tfrac{2}{9}(0.5) + \tfrac{3}{9}(0.8) \\ &= \frac{1.2 + 1.0 + 2.4}{9} = \frac{4.6}{9} \\ &= \frac{46}{90} = \frac{23}{45} \end{aligned} \]
The other fractions come from wrong denominators: 55 does not arise from the 4 + 2 + 3 = 9 split, and 32/45 is not the weighted sum. A quick check: the answer, about 0.51, must lie between the smallest and largest individual probabilities, 0.3 and 0.8.
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