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
Cheers,
Brent
