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29 Jun 2022, 01:30
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (02:56) correct
67%
(03:01)
wrong
based on 66
sessions
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Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let p be the probability that all four slips bear the same number. Let q be the probability that two of the slips bear a number a and the other two bear a number \(b \neq a\). What is the value of q/p?
(A) 162
(B) 180
(C) 324
(D) 360
(E) 720
Are You Up For the Challenge: 700 Level Questions
(A) 162
(B) 180
(C) 324
(D) 360
(E) 720
Are You Up For the Challenge: 700 Level Questions
29 Jun 2022, 05:48
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P:
There are 10 ways to choose which number shows up 4 times.
The number of ways to choose 4 slips are:
40!/4!36!
So P = 10/(40!/4!36!)
Q:
There are 10!/2!8! ways to select which 2 numbers.
There are 4!/2!2! ways to select 2 slips for each number
So Q = (4!/2!2!)^2*(10!/2!8!)/(40!/4!36!)
Q/P is therefore
(4!/2!2!)^2*(10!/2!8!)/10 =
36*45/10 = 162
Posted from my mobile device
There are 10 ways to choose which number shows up 4 times.
The number of ways to choose 4 slips are:
40!/4!36!
So P = 10/(40!/4!36!)
Q:
There are 10!/2!8! ways to select which 2 numbers.
There are 4!/2!2! ways to select 2 slips for each number
So Q = (4!/2!2!)^2*(10!/2!8!)/(40!/4!36!)
Q/P is therefore
(4!/2!2!)^2*(10!/2!8!)/10 =
36*45/10 = 162
Posted from my mobile device
29 Jun 2022, 19:06
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Bunuel
FWIW, this is a more complicated combinations question than any in the OG or OG Quant, so it's probably beyond the level of difficulty that you'd see on the actual test unless you're at Q50 or Q51.
If you're fluent in nCr and don't make mistakes in deploying that concept incorrectly, that's probably best on this question. If you're not fluent in that and are still in the Q49+ range, here's another way to think it through.
P: The odds of drawing SOME number on the first slip is 1. The odds of drawing the same number on the second slip is 3/39. The odds of drawing the same number on the third slip is 2/38. The odds of drawing the same number on the fourth slip is 1/37. So, the odds of doing all four of those things is P = (3*2*1)/(39*38*37).
Q: There are only two ways to get there. Case 1: We could draw SOME number with the first two slips and then a DIFFERENT number with the next two slips. Case 2: we could draw two different numbers with the first slip and second slip and then draw one more of each with slips three and four. Let's look at those cases separately.
What are the odds of drawing some number the first two slips and then a different number with slips three and four? 1*(3/39)*(36/38)*(3/37). That's (3*36*3)/(39*38*37)
What are the odds of drawing some number first, a different number second, and then one more of each? 1*(36/39)*(6/38)*(3/37). That's (36*6*3)/(39*38*37)
Q is case1 + case2 = (9*36*3)/(39*38*37)
Q/P = (9*36*3)/(3*2*1) = 9*6*3 = 54*3 = 162
Answer choice A.
