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Dime Issue

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by vishugogo » Sun Apr 14, 2013 6:08 am
Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. If she withdraws a number of these coins at random, how many coins would she have to withdraw to ensure that she has at least a 50 percent chance of withdrawing at least one quarter?

A) 1
B) 2
C) 5
D) 6
E) 7

No idea how to approach this problem....
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by Brent@GMATPrepNow » Sun Apr 14, 2013 6:28 am
vishugogo wrote:Karen has exactly 6 quarters, 5 dimes, and 10 nickels in her pocket. If she withdraws a number of these coins at random, how many coins would she have to withdraw to ensure that she has at least a 50 percent chance of withdrawing at least one quarter?

A) 1
B) 2
C) 5
D) 6
E) 7

No idea how to approach this problem....
I suggest a combination of gut feelings and checking the answer choices.

First off, there are 21 coins altogether and 6 are quarters. So, we can definitely see that Karen must select more than 1 coin so that P(select a quarter is > 0.5
So, we'll ELIMINATE answer choice A.

What about selecting 5 coins (answer choice C)? Well, that seems like a lot more than are necessary (gut feeling here)

So, let's test answer choice B (select 2 coins)

We want P(select at least 1 quarter)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 quarter) = 1 - P(not getting at least 1 quarter)
What does it mean to not get at least 1 quarter? It means getting zero quarters.
So, we can write: P(getting at least 1 quarter) = 1 - P(getting zero quarters)

Now let's calculate P(getting zero quarters)
What needs to happen in order to get zero quarters?
Well, we need a non-quarter for the first selection AND a non-quarter for the second selection.
We can write P(getting zero quarters) = P(non-quarter on 1st AND non-quarter on 2nd)
This means that P(getting zero quarters) = P(non-quarter on 1st) x P(non-quarter on 2nd)
= (15/21) x (14/20)
= (5/7) x (7/10)
= 1/2

So P(getting at least 1 quarter ) = 1 - P(getting zero quarters)
= 1 - 1/2
= 1/2

So, when Karen selects 2 coins, the probability is 1/2 that she selects at least 1 quarter.
Answer = B

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Brent
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by sana.noor » Thu Aug 01, 2013 10:50 am
if Karen select 7 coins then chance to select one quarter will be 1/7= 14%
chance to select quarter from 6 coins= 1/6 = 16%
chance to select one quarter from 5 coins = 1/5 = 20%
chance to select one quarter from 4 coins = 1/4 = 25%
chance to select one quarter from 3 coins = 1/3 = 33%
chance to select one quarter from 2 coins = 1/2 = 50%

So B is the answer
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by Matt@VeritasPrep » Thu Aug 01, 2013 4:34 pm
Maybe an easier way of doing it:

(Odds of AT LEAST ONE) = 1 - (Odds of NONE)

So we want

1 - (Odds of No Quarters) ≥ 1/2

For one coin, we have
(Odds of At Least One Quarter)
= 1 - (Odds of No Quarters)
= 1 - 15/21
= 6/21


For two coins, however, we have

(Odds of At Least One Quarter)
= 1 - (Odds of No Quarters)
= 1 - (15/21)(14/20)
= 1 - (2/3)(15/20)
= 1 - 10/20
= 1/2

So two coins.
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by vipulgoyal » Mon Aug 05, 2013 1:46 am
Alt way

6 quarters, 5 dimes and 10 nickels. If she were to pick 1 coin, the probability of it being a quarter is 6 / 21 ~ .28
qustion is asking "how many coins would she have to withdraw"if it would be the least no of coins then the ans would have been stright away 1"
Assume the first coin was a dime or a nickel. Now she picks one more coin and the resulting probability for a quarter is 6 / 20 = .3

Since they are discrete events, adding the 2 probabilities, yields 0.28 + 0.3 which is a little more than 50%.
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