1. probability

There are one blue, two black and three yellow balls. Two balls are randomly chosen without replacement. Define the following events: One of the balls is blue
At least one ball is black
Both balls are yellow
Both balls are of the same color
Find the following conditional probabilities:
(a)
(b)
(c) =

how can i solve this??

2. (a) 0 (there are no black balls)
(b) 0 (there are no black balls)
(c) = 0 (if one is blue then both can not be yellow)

Unless I misunderstand the question=)

3. ive updated it, now my question is correct, so how can i solve this

4. Originally Posted by makaveli89
ive updated it, now my question is correct, so how can i solve this
Originally Posted by makaveli89
There are one blue, two black and three yellow balls. Two balls are randomly chosen without replacement. Define the following events: One of the balls is blue

At least one ball is black

Both balls are yellow

Both balls are of the same color

Find the following conditional probabilities:

(a)

(b)

(c) =

how can i solve this??
The first thing you should do is draw a tree diagram. Have you done this?

Then it's much easier to see that:

$\Pr(B | A) = \frac{\Pr(B \cap A)}{\Pr(A)} = \frac{\Pr(\text{Black ball and blue ball})}{\Pr(A)}$ $= \frac{\left(\frac{1}{6}\right) \cdot \left(\frac{2}{5}\right) + \left(\frac{2}{6}\right) \cdot \left(\frac{1}{5}\right)}{\frac{1}{6} + \left(\frac{2}{6}\right) \cdot \left(\frac{1}{5}\right) + \left(\frac{3}{6}\right) \cdot \left(\frac{1}{5}\right)} = \, ....$.

$\Pr(D | B) = \frac{\Pr(D \cap B)}{\Pr(B)} = \frac{\Pr(\text{Both balls are black})}{\Pr(B)} = \frac{ \left(\frac{2}{6}\right) \cdot \left(\frac{1}{5}\right) }{\left(\frac{1}{6}\right) \cdot \left(\frac{2}{5}\right) + \frac{2}{6} + \left(\frac{3}{6}\right) \cdot \left(\frac{2}{5}\right) } = \, ....$

$Pr(C | A) = 0$.

5. wow thank you its correct. but where did u get all the numbers from?

6. Originally Posted by makaveli89
wow thank you its correct. but where did u get all the numbers from?
Draw the tree diagram like I suggested. It will help you see where the numbers came from.