M14-01

I'm confused. Can someone help me here?

Probability = Favourable Outcomes/Total possible outcomes

In this case, since the first pick is not being replaced, shouldn't the Probability of the second pick be 5/7+4/7 ?

Official Solution:

A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?

A. \(\frac{1}{3}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{8}\)
E. \(\frac{11}{16}\)


There are 2 different scenarios to evaluate:

1. Mary will extract a white ball, John will extract a blue ball. The probability of this is \(\frac{3}{8} * \frac{5}{7} = \frac{15}{56}\).

2. Mary will extract a blue ball, John will extract a blue ball. The probability of this is \(\frac{5}{8} * \frac{4}{7} = \frac{20}{56}\).

The overall probability that John will extract a blue ball = \(\frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}\).

Alternative Explanation

The initial probability of drawing blue ball is \(\frac{5}{8}\). Without knowing the other results, the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still would be \(\frac{5}{8}\).


Answer: D

Hi Bunnuel,

Could you please elaborate a bit on the alternative solution? Why would 1st draw not change the probability of the second draw as the set from where the second draw is made would consist just of 7 balls and we aren't sure on the count of BLUE and WHITE in this draw as we do not know the result of the first.

Did it by the first process but took me close to 3 mins.

Best-
Amit