A Basket Contains 3 White and 5 Blue Balls GMAT Problem Solving

Question: A basket contains 3 white and 5 blue balls. Marry 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?

1. \(\frac{1}{3}\)
2. \(\frac{3}{8}\)
3. \(\frac{1}{2}\)
4. \(\frac{5}{8}\)
5. \(\frac{11}{16}\)

“A basket contains 3 white and 5 blue balls. Marry 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?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1

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 \(\frac{15}{56}+\frac{20}{56}=\frac{35}{56}=\frac{5}{8}\)

Correct Answer: D

Approach Solution 2

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 I 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 \(\frac{5}{8}\) .

Correct Answer: D

Approach Solution 3

1. Possible arrangements of 8 marbles in a row with \(2^{nd}\) marble is blue = \(\frac{7!}{3!*4!}\)

Next step to calculate the probability:

2. Possible arrangement of 8 marbles without any restriction = \(\frac{8!}{3!*5!}\)

Probability = \(\frac{(1)}{(2)}=\frac{5}{8}\)

Correct Answer: D

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