Question

In a game of throwing two dice, one may receive

₹ 30

If none of the dice shows a

5

or a

6

. But he has to pay

₹ 10

if at least one die shows a

5

or a

6

. In another game of tossing three coins, one may receive

₹ 80

if all heads or all tails are shown. But, in this game he has to pay

₹ 30

if only one or only two heads appear. A person has enough money to play any of these two games. Which one will he prefer to play?

Accepted Answer

Let $X$ denote the amount received by the person if he plays the first game. Then the values that X can take are $30$ and $- 10$ .

Now, $5$ The probability that none of the dice shows a $₹ 30$ or a $6$

$= \frac{4}{6} \times \frac{4}{6} = \frac{4}{9}$

$P (X = - 10) =$ The probability that at least one die shows is $1 - \frac{4}{9} = \frac{5}{9}$

Thus, the expectation of the random variable $X = E (X)$

$= 30 \times \frac{4}{9} - 10 \times \frac{5}{9}$

$= \frac{120 - 50}{9} = \frac{70}{9} = 7.78$ (approx.)
So, in case of this game, a gain of $₹ 7.78$ is expected.

Suppose that $Y$ denotes the amount received by the person if he plays the second game. Then $Y$ takes the values $80$ and $- 30$

Sample space $S = \{H H H, T T T, H T T, T H T, T T H, T H H, H T H, H H T\}$

Now, $P (Y = 80) =$ The probability of getting all heads or all tails $= \frac{2}{8} = \frac{1}{4}$

$P (Y = - 30) =$ The probability that only one or only two heads are shown

$= \frac{6}{8} = \frac{3}{4}$

Thus, the expectation of the random variable $Y = E (Y)$

$= 80 \times \frac{1}{4} - 30 \times \frac{3}{4} = \frac{- 10}{4} = - 2.5$

So, in case of the second game, a loss of $₹ 2.50$ is expected

Hence, the person will prefer the first game to play.

Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 4: EXERCISE

Parthasarathi Mukhopadhyay Mathematics Solutions for Exercise - Parthasarathi Mukhopadhyay and Manabendra Nath Mukherjee Solutions for Exercise 4: EXERCISE

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