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A bag contains $3$ red, $6$ yellow and $7$ blue balls. What is the probability that the two balls drawn are yellow and blue ?

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up vote 3 down vote accepted

If you do not replace the balls after they are drawn (I'm assuming this is your problem):

Thinking of the balls as being distinct and imagining that we draw one ball, and then the other, there are $16\cdot15$ possible (ordered) outcomes.

The number of outcomes where one ball is yellow and the other blue is $$ \underbrace{6\cdot 7}_{\text{ yellow first }}+\underbrace{7\cdot 6}_{\text{ blue first }}=84. $$ Since outcomes where we record the color of the drawn balls in order are equally likely, the probability that one ball is yellow and the other blue is $$ {\text{number of desired outcomes}\over\text {total number of outcomes}}={84\over 16\cdot 15}={ 21\over 4\cdot15}={7\over 20}. $$

If the balls are replaced after drawing, then the total number of outcomes is $16\cdot16$ and the number of outcomes in which one is yellow and the other blue is $2\cdot 7\cdot 6$. The probability that one is yellow and the other blue is ${2\cdot7\cdot6\over 16\cdot16}= {21\over64}.$

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Excellent explanation. Thanks David. – Nikhil Mulley Dec 13 '11 at 13:37
I am sure I can learn lot of math here on this site from here on .. :-) – Nikhil Mulley Dec 13 '11 at 13:37
@Nikhil Thanks. I'm glad to help :) – David Mitra Dec 13 '11 at 13:39

$$P(\mbox{blue, yellow})=P(\mbox{1st is blue, 2nd is yellow})+P(\mbox{1st is yellow, 2nd one is blue})$$ $$=\frac{7}{16}\cdot\frac{6}{15}+\frac{6}{16}\cdot\frac{7}{15}=\frac{7}{20}.$$

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Please note there are $6$ yellow balls. – Quixotic Dec 13 '11 at 13:38
@MaX: hahaha! Yes, you are right. Edited now. – Paul Dec 13 '11 at 13:39
$+1$ for you :-) – Quixotic Dec 13 '11 at 13:41

Simplest way: There are ${7 \choose 1}$ ways of choosing a blue ball, ${6 \choose 1}$ ways of choosing a yellow ball and ${16 \choose 2}$ ways of choosing two balls at random out of 16.

Required probability is therefore $ \frac {{7 \choose 1}{6 \choose 1}}{16 \choose 2}$

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