Problem Solving

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

Hi
IMO the answer is B

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

thanks and regards

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?

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

You started out really well but then missed the execution. See Image below

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).

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)

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