In roulette, Figure, the wheel has 13 numbers 0, 1, 2, ...., 12 marked on equally spaced slots. A player sets Rs 10 on a given number. He receives Rs 100 from the organiser of the game if the ball comes to rest in this slot; otherwise he gets nothing. If X denotes the player's net gain/loss, find E (X).
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Solution

The wheel has 13 numbers, i.e. 0, 1, 2, ... ,12 marked on equally spaced slots.

∴ Probability of ball resting on any particular number = $\frac{1}{13}$

Let the player set Rs 10 on a given number k.

P(player receives Rs 100) = P(ball rests on it) = $\frac{1}{13}$

X denotes the player's net gain or loss. If he gets the required number, then his gain is Rs 90 (100-10).

If the ball does not rest on the number, then it rests on any of the other 12 numbers. In that case, the player's loss is Rs 10.

Thus, the probability distribution of X is given by

Computation of mean

x_i p_i p_ix_i

90 $\frac{1}{13}$ $\frac{90}{13}$

$-$ 10 $\frac{12}{13}$ $\frac{- 120}{13}$

$\sum_{}^{}$ p_ix_i = $\frac{- 30}{13}$

$Mean = E (x) = \sum p_{i} x_{i} = - \frac{30}{13}$

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x_i	p_i	p_ix_i
90	$\frac{1}{13}$	$\frac{90}{13}$
$-$ 10	$\frac{12}{13}$	$\frac{- 120}{13}$
		$\sum_{}^{}$ p_ix_i = $\frac{- 30}{13}$