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A bag has 30 different coins, of which 20 are fair. 2 coins
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18 Sep 2019, 13:43
18 Sep 2019, 13:43
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Question Stats:
44% (01:34) correct
55% (01:45) wrong based on 92 sessions
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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
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19 Sep 2019, 03:26
19 Sep 2019, 03:26
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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)
