Roland2rule wrote:Jean drew a gumball at random from a jar of pink and blue gumballs. Since the gumball she selected was blue and she wanted a pink one, she replaced it and drew another. The second gumball also happened to be blue and she replaced it as well. If the probability of her drawing the two blue gumballs was 9/49, what is the probability that the next one she draws will be pink?

A. 1/49
B. 4/7
C. 3/7
D. 16/49
E. 40/49

OA is B
I couldn't get the word interpretation of this question.Can an expert do justice pls? Thanks

Say there are x numbers of blue and y numbers of pink balls.

Thus, the probability of drawing a blue ball = x/(x + y) and the probability of drawing two blue balls (one after the other) = [x/(x + y)]^2 = x^2/(x+y)^2

Thus, we have x^2/(x + y)^2 = 9/49 = 3^2/7^2 = 3^2/(3 + 4)^2

=> x = 3p and y = 4p; where p is a positive integer

The probability that the next one she draws will be pink = Number of pink balls / Total number of balls = 4p / (3p + 4p) = 4p/7p = 4/7.

The correct answer: B

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: Pasadena | Warsaw | Cairo | Kuala Lumpur | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

BTGmoderatorRO wrote:Jean drew a gumball at random from a jar of pink and blue gumballs. Since the gumball she selected was blue and she wanted a pink one, she replaced it and drew another. The second gumball also happened to be blue and she replaced it as well. If the probability of her drawing the two blue gumballs was 9/49, what is the probability that the next one she draws will be pink?

A. 1/49
B. 4/7
C. 3/7
D. 16/49
E. 40/49

Let the probability of drawing a blue gumball be p. Then the probability of drawing two blue gumballs with replacement is 9/49, which means that p^2 = 9/49. Therefore, we have p = √(9/49) = 3/7.

Since the probability of drawing a pink gumball is (1 - p), the probability of drawing a pink gumball is 1 - 3/7 = 4/7.

Answer: B