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Could anyone solve the following with a combinatorial approa

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Could anyone solve the following with a combinatorial approa

by BTGmoderatorAT » Sun Jan 07, 2018 9:13 am
Could anyone solve the following with a combinatorial approach instead of probability one?

Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?

a) 8/125.
b) 1/30.
c) 2/5.
d) 1/720.
e) 3/10.

I'm confused how to set up the formulas here. Can any experts help?
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by Brent@GMATPrepNow » Sun Jan 07, 2018 10:13 am
ardz24 wrote:Could anyone solve the following with a combinatorial approach instead of probability one?

Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?

a) 8/125.
b) 1/30.
c) 2/5.
d) 1/720.
e) 3/10.
Here's the counting approach:

P(all 3 mice are black) = (# of ways to select 3 black mice)/(TOTAL # of ways to select ANY 3 mice)

# of ways to select 3 black mice
Since the order in which we select the mice does not matter, we can use COMBINATIONS
We can select 3 black mice from 4 black mice in 4C3 ways (= 4 ways)

TOTAL # of ways to select ANY 3 mice
We can select 3 mice from all 10 mice in 10C3 ways (= 120 ways)

ASIDE: If anyone is interested, we have a video on calculating combinations in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

So, P(all 3 mice are black) = 4/120
= 1/30
= B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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