A bag contains 5 red and 7 black balls. Second bag contains 4 blue and 3 green balls. 1 ball is drawn from each bag. Find the probabilty for 1 red and 1 blue ball.
The answer is 5/21 But don't know the way to get it. Plzz help.
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$\begingroup$ $\frac{5}{12}\cdot\frac{4}{7}$ should be the answer. $\endgroup$ – Poppy Jan 20 '14 at 7:44
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Probability of getting a red ball from the first bag - $\frac{5}{12}$.
Probability of getting a blue ball from the second bag - $\frac{4}{7}$.
Since they are independent the probability of having both is a multiplication of the probability of each - $\frac{(5 * 4)}{(7 * 12)} = \frac{20}{84} = \frac{5}{21}$.
Enjoy...
