Question

Mangoes numbered 1 through 18 are placed in a bag for delivery. Two mangoes are drawn out of the bag without replacement. Find the probability such that all the mangoes have even numbers on them?

(a) 43.7%

(b) 34%

(c) 6.8%

(d) 9.3%

This question was addressed to me by my school teacher while I was bunking the class.

The doubt is from Discrete Probability topic in portion Discrete Probability of Discrete Mathematics

Answer 1

Correct choice is (c) 6.8%

Best explanation: The events are not independent. There will be a \(\frac{10}{18} = \frac{5}{9}\) chance that any of the mangoes in the bag is even. The probability that the first one is even is \(\frac{1}{2}\), for the second mango, given that the first one was even, there are only 9 even numbered balls that could be drawn from a total of 17 balls, so the probability is \(\frac{9}{17}\). For the third mango, since the first two are both odd, there are 8 even numbered mangoes that could be drawn from a total of 16 remaining balls and so the probability is \(\frac{8}{16}\) and for fourth mango, the probability is = \(\frac{7}{15}\). So the probability that all 4 mangoes are even numbered is \(\frac{10}{18}*\frac{9}{17}*\frac{8}{16}*\frac{7}{16}\) = 0.068 or 6.8%.