Question

Let X = {1, 2, 3}. A number is selected from set X with replacement and this process is done 1100 times. For each i, $1 \leq i \leq 3$ . Let f (i) represent the number of times the number i is selected. Clearly f (1) + f (2) + f (3) = 1100.

Also, let S denote the total sum of 1100 numbers selected. If S³ = 162 f (1) f (2) f (3).

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Linked Question 1

The value of harmonic mean of f (1), f (2) and f (3) is

400

100

300

200

$∵$ S = f (1) + 2f (2) + 3 f (3)

$∵$ AM $\geq$ GM $\Rightarrow \frac{S}{3} \geq$ (6f (1) f (2) f (3))^1/3

⇒ S³ $\geq$ 162 f (1) f (2) f (3)

but given S³ = 162 f (1) f (2) f (3)

$∴$ f (1) = 2 f (2) = 3 f (3)

& f (1) + f (2) + f (3) = 1100

$∴$ f (1) = 600, f (2) = 300 and f (3) = 200

$∴ HM = \frac{3}{\frac{1}{f (1)} + \frac{1}{f (2)} + \frac{1}{f (3)}}$

$= \frac{300}{\frac{1}{6} + \frac{1}{3} + \frac{1}{2}} = 300$

Linked Question 2

There are f (1) number of red balls, f (2) number of blue balls and f (3) number of white balls in a bag. Two persons A & B in order draws a ball from the bag & put it back after noting its colour. Person who draws a white ball first wins the game. If A starts the game then probability of winning of B is

$\frac{11}{20}$

$\frac{9}{20}$

$\frac{2}{11}$

$\frac{9}{11}$

(ii) Probability of white ball = $\frac{2}{11}$

& probability of not getting white ball = $\frac{9}{11}$

$∴$ probability of winning of

$B = \frac{9}{11} \times \frac{2}{11} + \frac{9}{11} + \frac{9}{11} + \frac{9}{11} \times \frac{2}{11} + . ...........$

$= \frac{\frac{9}{11} \times \frac{2}{11}}{1 - \frac{9}{11} \times \frac{9}{11}} = \frac{18}{40} = \frac{9}{20} .$