Conditional Probability Question

**RedFive** · February 6th 2010, 02:56 PM

I was wondering if I could get some help with a problem. I have thought about this for a long time and can't figure it out. I don't need to solve the problems (at least not yet), I just need to know if they have different answers and if so, why. I thought they should all have the same answer, but my calculations are not working out that way :-(

Imagine a glass with 6 white balls numbered 1-6 and 4 black balls numbered 1-4. 3 balls are chosen at random.

A) What is the probability you choose 3 white balls given you choose at least 1 white ball.

B) What is the probability you choose 3 white balls given you choose at least 1 even numbered white ball?

C) What is the probability you choose 3 white balls given you choose white ball #2?

**Random Variable** · February 6th 2010, 03:59 PM

I'm not entirely confident with my answers.

A) $\frac{\binom{6}{3} \binom{4}{0}}{\binom{6}{1} \binom{4}{2} +\ \binom{6}{2} \binom{4}{1} + \binom{6}{3} \binom{4}{0}}$

B) $\frac{\binom{3}{1} \binom{5}{2} \binom{4}{0}}{\binom{3}{1} \binom{5}{2} \binom{4}{0} + \binom{3}{1} \binom{5}{1} \binom{4}{1} + \binom{3}{1} \binom{5}{0} \binom{4}{2}}$ because you want to choose at least one of the 3 even numbered white balls

C) $\frac{1 \binom{5}{2} \binom{4}{0}}{1 \binom{5}{2} \binom{4}{0} + 1 \binom{5}{1} \binom{4}{1} + 1 \binom{5}{0} \binom{4}{2}}$

**matheagle** · February 6th 2010, 10:31 PM

I don't trust (b), since you only have 10 balls, the sum of the number of balls doesn't add up to 10 here.

**Random Variable** · February 7th 2010, 03:02 AM

Originally Posted by matheagle

I don't trust (b), since you only have 10 balls, the sum of the number of balls doesn't add up to 10 here.

Yeah, that answer doesn't make any sense.

By splitting the white balls into two groups, I get the following answer which makes a bit more sense:

$\frac{\binom{3}{1} \binom{3}{2} \binom{4}{0} + \binom{3}{2} \<br /> \binom{3}{1} \binom{4}{0} + \binom{3}{3}\binom{3}{0}\binom{4}{0}}{\binom{3}{1} \binom{3}{2} \binom{4}{0} + \binom{3}{1} \binom{3}{1} \binom{4}{1} + \binom{3}{1} \binom{3}{0} \binom{4}{2} + \binom{3}{2} \binom{3}{1} \binom{4}{0} + \binom{3}{2} \binom{3}{0} \binom{4}{1} + \binom{3}{3} \binom{3}{0} \binom{4}{0}}$

**matheagle** · February 7th 2010, 07:12 AM

I get the same denominator, but I would only consider TWO groups.
I lump the odd whites with the black balls.

${3\choose 1}{7\choose 2}+{3\choose 2}{7\choose 1}+{3\choose 3}{7\choose 0}=85$

Thread: Conditional Probability Question

LinkBack

Thread Tools

Display

Conditional Probability Question

Similar Math Help Forum Discussions

Conditional Probability and Conditional Expectation Question

Conditional probability question help

Conditional Probability question

Conditional probability question

Conditional probability question

Search Tags