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Posts: 17
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Julia and Brian play a game in which Julia takes a ball and if it is green, she wins. If the first ball is not green, she takes the second ball (without replacing first) and she wins if the two balls are white or if the first ball is gray and the second ball is white. What is the probability of Julia winning if the jar contains 1 gray, 2 white and 4 green balls?
A. \(\frac{2}{7}\)
B. \(\frac{4}{7}\)
C.\(\frac{1}{7}\)
D. \(\frac{2}{6}\)
E. \(\frac{2}{3}\)
A. \(\frac{2}{7}\)
B. \(\frac{4}{7}\)
C.\(\frac{1}{7}\)
D. \(\frac{2}{6}\)
E. \(\frac{2}{3}\)
IanStewart
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Joined: 24 Jun 2008
Last visit: 28 Aug 2026
Posts: 4,146
Location: United States
GMAT 1: 780 Q51 V47
Expert reply
GMAT 1: 780 Q51 V47
Posts: 4,146
She wins by picking a green marble, so she wins 4/7 of the time that way. But she also can win in other ways, so the answer must be greater than 4/7. Only E could be right.
Saurabjhain296 posted a perfect solution above using cases, which is how I'd do the problem too if the answer choices weren't so convenient.
_________________
Saurabjhain296 posted a perfect solution above using cases, which is how I'd do the problem too if the answer choices weren't so convenient.
_________________
now teaching gmat focus
ianstewartgmat.com
ianstewartgmat.com
General Discussion
No of ways Julia can win:
I. If the first ball drawn by Julia is Green- Probability (P1) = No of favorable outcomes / Total outcomes = 4/7
II. If the first ball drawn by Julia is not green and first & the second ball drawn by Julia are White- Probability (P2) = (2/7)*(1/6) = 1/21
III. If the first ball drawn by Julia is gray and the second is white- Probability (P3) = (1/7)*(2/6) = 1/21
Total probability of win = P1 + P2 + P3 = (4/7) + (1/21) + (1/21) = 14/21 = 2/3
I. If the first ball drawn by Julia is Green- Probability (P1) = No of favorable outcomes / Total outcomes = 4/7
II. If the first ball drawn by Julia is not green and first & the second ball drawn by Julia are White- Probability (P2) = (2/7)*(1/6) = 1/21
III. If the first ball drawn by Julia is gray and the second is white- Probability (P3) = (1/7)*(2/6) = 1/21
Total probability of win = P1 + P2 + P3 = (4/7) + (1/21) + (1/21) = 14/21 = 2/3
Posts: 17
saurabhjain296
Hi,
Can you please explain the 1/6 in II and 2/6 part in III?
Posts: 6
Aprajita760
Probability of winning will be all probabilities combined. Hence:
Probability of picking green (1st ball), also probability of winning (P1)= 4/7
Now, Probability of picking white ball in first attempt= 2/7
Probability of picking another white ball in 2nd attempt= 1/6 (It is 1/6 because one white ball has been already picked from the 2 and 1 ball from total outcomes hence 6)
Probability of winning (P2)= 2/7*1/6= 1/21
Probability of picking 1st ball as grey= 1/7
Probability of picking 2nd ball as white= 2/6 (2 because we have total of 2 white balls out of which none have been picked yet & 6 because one ball is picked out of the total)
Probability of winning (P3)= 1/7*2/6= 1/21
Total probabilty of winning= P1 + P2 + P3
ashmitakhan
But since she goes on to pick the second ball(if the first ball is not green) shouldn't the number of balls reduce by 1 i.e 6?
This is a good question to understand how and & or works.
Here, there are 3 OR situations
1. If first ball is green
2. If first ball is not green, then she takes another ball(without replacing) and she wins if both the balls are white
3. If the first ball is gray and second ball is white.
When ever we have OR, we have to add & when ever we have AND, we have to multiply
Now, let's take the situations.
So, there are 7 balls.
Probability of getting 1st ball as green = \(\frac{4}{7}\) = P(1)
Note :-After taking one ball, that ball is kept with her because there is no replacement.
So, remaining 6 balls. We have to apply this when more balls are selected.
OR
Probability of getting white balls. Here we can see how AND comes into play.
= \(\frac{2}{7}\)*\(\frac{1}{6} \)= \(\frac{1}{21}\) = P(2)
OR
Probability of getting gray as first ball and white as second.
= \(\frac{1}{7}\)*\(\frac{2}{6}\) = 1/21 = P(3)
Now, overall Probability = P(1)+P(2)+P(3)
= \(\frac{4}{7}\)+\(\frac{1}{21}\)+\(\frac{1}{21}\) = \(\frac{2}{3}\)
Answer E
Here, there are 3 OR situations
1. If first ball is green
2. If first ball is not green, then she takes another ball(without replacing) and she wins if both the balls are white
3. If the first ball is gray and second ball is white.
When ever we have OR, we have to add & when ever we have AND, we have to multiply
Now, let's take the situations.
So, there are 7 balls.
Probability of getting 1st ball as green = \(\frac{4}{7}\) = P(1)
Note :-After taking one ball, that ball is kept with her because there is no replacement.
So, remaining 6 balls. We have to apply this when more balls are selected.
OR
Probability of getting white balls. Here we can see how AND comes into play.
= \(\frac{2}{7}\)*\(\frac{1}{6} \)= \(\frac{1}{21}\) = P(2)
OR
Probability of getting gray as first ball and white as second.
= \(\frac{1}{7}\)*\(\frac{2}{6}\) = 1/21 = P(3)
Now, overall Probability = P(1)+P(2)+P(3)
= \(\frac{4}{7}\)+\(\frac{1}{21}\)+\(\frac{1}{21}\) = \(\frac{2}{3}\)
Answer E
(4c1+2c2+1c1*2c1)/7c2=1/3. what is wrong here , anyone?
Thank you!! This is a more detailed response. I always used to get confused when we should add or multiply conditional probabilities.
Sr1994
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