Three balls are marked 1,2 and 3. They are placed in a bowl and a ball is drawn, its number is recorded and the ball is returned to the bowl. The process is repeated two more times. If the sum of the three numbers is 6 then $P = \frac{a}{b}$ is the probability (express in lowest form) that the ball numbered 2 was drawn all the three times, then find the value of $(a + b) .$

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answer is 8.

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Detailed Solution

   $H_{1} :$ Three numbers drawn are 1, 2 and 3 in any order.
$H_{2} :$ Three numbers drawn are 1, 2 and 2. $P (H_{1}) = 6 (\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}) = \frac{2}{9}$
think! $P (H_{2}) = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{27}$
   $∴ P (H_{2} / H_{1} \cup H_{2}) = \frac{P (H_{2})}{P (H_{1}) + P (H_{2})} = \frac{1}{27} \cdot \frac{27}{7} = \frac{1}{7} \equiv \frac{a}{b}$ $∴ (a + b) = 8.$

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