The player wins $6, that is, Ayo loses £6, if he rolls a 6 and spins a 1, hence the probability is $\frac{1}{6} \times \frac{1}{12} = \frac{1}{72}$ .

The player wins $3, that is, Ayo loses £3, if he rolls a 3 on at least one of the spinner or the dice, hence, considering three cases(both and either the spinner of the dice), the probability is $\frac{1}{6} \times \frac{1}{12} + \frac{1}{6} \times \frac{11}{12} + \frac{5}{6} \times \frac{1}{12} = \frac{1 + 11 + 5}{72} = \frac{17}{72}$

In the other cases, Ayo wins £1.40, with $1 - \frac{18}{72} = \frac{54}{72}$ probability.

Hence, his expected profit for a single game is:

$E(X) = -6\frac{1}{72} - 3\frac{17}{72} + 1.4\frac{54}{72} = \frac{-6 - 3(17) + 54(1.4)}{72} = 0.2583$

For 216 games, the expected value is:

$E = 216(0.2583) = 55.8$

Ayo can be expected to make a profit of £55.8.

To learn more about expected value, you can take a look at brainly.com/question/24855677

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2 years ago