PUNE MCA Exam 2017 Question Paper with answers for online practice

Concept:Convert the given odds into probabilities, then compute the probability that at least two critics favour the book.Explanation:For each critic, probability of favour = $\frac{\text{favourable odds}}{\text{favourable} + \text{unfavourable}}$.Critic 1: odds 5:2? Wait, let's use the common denominator 7. If the first critic's odds were 5:2, probability $\frac{5}{7}$; but given options suggest $\frac{5}{7}$, $\frac{4}{7}$, $\frac{3}{7}$ as probabilities. So, using these:$P_1 = \frac{5}{7}$, $P_2 = \frac{4}{7}$, $P_3 = \frac{3}{7}$.Probability that exactly three favour: $\frac{5}{7} \times \frac{4}{7} \times \frac{3}{7} = \frac{60}{343}$.Probability that exactly two favour (three cases):Case (1,2 favour, 3 against): $\frac{5}{7} \times \frac{4}{7} \times \left(1 - \frac{3}{7}\right) = \frac{5}{7} \times \frac{4}{7} \times \frac{4}{7} = \frac{80}{343}$.Case (1,3 favour, 2 against): $\frac{5}{7} \times \frac{3}{7} \times \left(1 - \frac{4}{7}\right) = \frac{5}{7} \times \frac{3}{7} \times \frac{3}{7} = \frac{45}{343}$.Case (2,3 favour, 1 against): $\left(1 - \frac{5}{7}\right) \times \frac{4}{7} \times \frac{3}{7} = \frac{2}{7} \times \frac{4}{7} \times \frac{3}{7} = \frac{24}{343}$.Total for exactly two: $\frac{80+45+24}{343} = \frac{149}{343}$.Probability of majority (at least two): $\frac{60}{343} + \frac{149}{343} = \frac{209}{343}$.Answer:$\frac{209}{343}$