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A bit confused

by Nijo » Tue Jul 15, 2014 11:25 pm
Hi
Pl see the question below:
8. 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 tried solving it by 2 methods: In method 1, I chose by 4C3/ 10C3 and got the correct answer.
In method 2, I tried to figure prob(1st black ball) * prob (2nd ball is black) * prob(3rd ball is black)
That means 2/5 * (6/10* 4/9) * (6/10*5/9*4/8) but not getting the answer
I know its a simple question but I have tried n getting confused
Thanks

Regds
Nidhi
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by [email protected] » Tue Jul 15, 2014 11:59 pm
Hi Nijo,

You can approach this question in a couple of different ways. If you're going to treat it as a straight-forward probability question, then here's how you would do the math:

4 Black
6 White
10 Total

Probability of choosing 3 black mice is...

(4/10)(3/9)(2/8) = 1/30

Final Answer: B

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by unknown13 » Wed Jul 16, 2014 12:26 am
Hi
IMO the answer is B

used the method of combination
4C3/10C3
= 3!*4!*7!/3!*10!
= 1/30

thanks and regards
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by Nijo » Wed Jul 16, 2014 1:31 am
Thanks, that is the method I used to
I am not however getting the answer in Method 2 I used, any clue what am I doing wrong?
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by mcdesty » Wed Jul 16, 2014 3:28 am
In method 2, I tried to figure prob(1st black ball) * prob (2nd ball is black) * prob(3rd ball is black)
That means 2/5 * (6/10* 4/9) * (6/10*5/9*4/8) but not getting the answer
I know its a simple question but I have tried n getting confused
Thanks

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by [email protected] » Wed Jul 16, 2014 9:02 am
Hi Nijo,

Can you explain your calculation - what does each part of that large calculation mean to you?

Ultimately, the reason why your second approach didn't get you the correct answer is that there's something incorrect about your logic (and thus the math was wrong). It's an easy enough fix though; I just need to know your thought process.

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by Matt@VeritasPrep » Wed Jul 16, 2014 10:29 am
Surprised that no one has mentioned that this question is unanswerable (as written): we need to know whether the mice are replaced or not.

Assuming the mice are chosen WITHOUT replacement, we'd have

Prob first black = 4/10 = 2/5
Prob second black = 3/9 = 1/3
Prob third black = 2/8 = 1/4

Since we need all three of these to hold, our probability is 2/5 * 1/3 * 1/4, or 1/30.

Assuming the mice are chosen WITH replacement, we'd have (4/10) * (4/10) * (4/10), or 8/125.

Since these are both answers, there's no way to know! Sloppy question, unfortunately.

Also, Nidhi, I think your second method is factoring WHITE MICE into the equation when it doesn't need to. Probability of all three black is Prob(Black 1st) * Prob(Black 2nd) * Prob(Black 3rd), but you've got a few white mice (6/10) and a few numbers that don't seem to represent white or black mice (4/9, 5/9).
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by Nijo » Thu Jul 17, 2014 7:03 am
Thanks everyone
I have understood my mistake. I was including in white mice where it was not needed
That is calculating prob (2nd mouse being black) = P(1st mouse is non black) * P(2nd mouse is black)
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by GMATinsight » Thu Jul 17, 2014 10:10 am
All three picks are dependent events for the desired outcome (all three black) therefore all three picks must be considered at once, one after the other and all of them must be black.

Probability = 4C3 / 10C3 = 1/30

OR

Probability = (4/10) x (3/9) x (2/8) = 1/30
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