GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
A bag has 30 different coins, of which 20 are fair. 2 coins
[#permalink]
18 Sep 2019, 13:43
18 Sep 2019, 13:43
2
Expert Reply
8
Bookmarks
00:00
Question Stats:
44% (01:31) correct
55% (01:40) wrong based on 104 sessions
Hide Show timer Statistics
A bag has 30 different coins, of which 20 are fair. 2 coins are selected with replacement from the bag.
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Probability of selecting exactly one fair coin |
\(\frac{20}{30} \times \frac{10}{29}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: A bag has 30 different coins, of which 20 are fair. 2 coins
[#permalink]
19 Sep 2019, 03:26
19 Sep 2019, 03:26
6
2
Bookmarks
Carcass wrote:
A bag has 30 different coins, of which 20 are fair. 2 coins are selected with replacement from the bag.
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
OA is not provided
Quantity A |
Quantity B |
Probability of selecting exactly one fair coin |
\(\frac{20}{30} \times \frac{10}{29}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
OA is not provided
Here,
there are 2 possibilities of selecting exactly one fair coin
Possibility 1:
One fair coin in first selection and Not a fair coin in second selection
Then the probability = \(\frac{20}{30} * \frac{10}{30} = \frac{2}{9}\)
Possibility 2:
Not a fair coin in first selection and a Fair coin in second selection
Then the probability = \(\frac{10}{30} * \frac{20}{30} = \frac{2}{9}\)
Total = \(\frac{2}{9} + \frac{2}{9} = \frac{4}{9}\)
Now QTY B :
\(\frac{20}{30} \times \frac{10}{29}\) = here the denominator will be 29 * 3 that is way big compared to the denominator of QTY A .
Therefore QTY A > QTY B
General Discussion
Re: A bag has 30 different coins, of which 20 are fair. 2 coins
[#permalink]
19 Sep 2019, 02:21
19 Sep 2019, 02:21
Qty B
as the Probability in Qty A : 20 / 30 * 10 / 30 (Given that there is Replacement of Balls ) ( For Fair Ball to be picked Prob = 20 / 30 & Unfair Ball = 10 / 30)
as the Probability in Qty A : 20 / 30 * 10 / 30 (Given that there is Replacement of Balls ) ( For Fair Ball to be picked Prob = 20 / 30 & Unfair Ball = 10 / 30)
