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Question:

Suppose there are 4 bags. Bag 1 contains 1 black and a² – 6a + 9 red balls, bag 2 contains 3 black and a² – 6a + 7 red balls, bag 3 contains 5 black and a² – 6a + 5 red balls and bag 4 contains 7 black and a² – 6a + 3 red balls. A ball is drawn at random from a randomly chosen bag. The maximum value of probability that the selected ball is black, is

Explanation:

Each bag has a² – 6a + 10 balls.

Bags 1, 2, 3 and 4 contain 1, 3, 5 and 7 black balls respectively.

Probability of selecting a black ball from a specific bag is

$\frac{n}{a^2-6a+9}$

where n is the number of black balls in that bag.

A bag is selected at random.

∴ Probability of selecting a particular bag = $\frac{1}{4}$

∴ Probability that the ball selected from that randomly chosen bag is black

= $\frac{1}{4}$$\left(\frac{1}{a^2-6a+10}\right)$ + $\frac{1}{4}$$\left(\frac{3}{a^2-6a+10}\right)$ + $\frac{1}{4}$$\left(\frac{5}{a^2-6a+10}\right)$ + $\frac{1}{4}$$\left(\frac{7}{a^2-6a+10}\right)$

=$\frac{1}{4}$$\left(\frac{16}{a^2-6a+10}\right)$ = $\left(\frac{4}{a^2-6a+10}\right)$

Hence, option (d).