Question:

A bag contains blue and red balls only.
If one ball is drawn, the probability of selecting a blue ball is 1/2.
If two balls are drawn randomly without replacement, the probability of selecting a blue and then a red ball is 5/18.
Two balls are drawn randomly without replacement, what is the probability of drawing a red ball, given that the first ball drawn was blue?

(A) 2/9
(B) 5/18
(C) 4/9
(D) ½
(E) 5/9

“A bag contains blue and red balls only” - is a topic of the GMAT Quantitative reasoning section of GMAT. To solve GMAT Problem Solving questions a student must have knowledge about a good number of qualitative skills. GMAT Quant section consists of 31 questions in total. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.

Solutions and Explanation

Approach Solution : 1

Only blue and red balls are in the bag. Let us assume them as b and r respectively

The likelihood of drawing a blue ball is 1/2 if only one ball is drawn.

So, b/(b+r) = 1/2 or b=r

The probability of choosing a blue and then a red ball when two balls are drawn at random without replacement is 5/18.

So, b/(b+r) * r/b+r−1 = 1/2 * r/(2r−1) =5/18.........

r/(2r−1) = 5/18 ∗ 2 = 5/9........9r = 5(2r−1) = 10r−5.....r = 5 = b

The probability of selecting one red ball from a pool of 5 red and 4 blue balls will therefore be,

5/(5+4) = 5/9

Correct Answer: (E)

Approach Solution : 2

There are only two colors of balls, they are blue and red.

There are equal numbers of red and blue balls if P(blue) = 1/2.

Probability of withdrawing two balls, first the blue ball and then the second, without replacing them is therefore , 1/2*({x)/2x-1} = 5/18.

As a result, x= 5.

Now, the likelihood that red balls will be withdrawn at a later time when blue balls are withdrawn first is 5/9.

Correct Answer: (E)

Approach Solution : 3

P(B) = 1/2

Two balls are drawn without replacements, which means P(a blue and a red ball) is 5/18.

This indicates that it is an independent even and, as a result P(B∩R) = P(B) * P(R)

According to Conditional Probability,

Chance of drawing a red ball given that the first ball drawn was blue will be

P(R/B) = [P(B∩R)] / P(B)

P(R/B) = (5/18)/0.5

P(R/B) = 5/9

Correct Answer: (E)

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