Question

a fruit machine consist of three windows, each of which shows pictures of fruits :lemon or oranges or cherries or plums. The probability that a window shows a particular fruit is as follows:
P(lemons)= 0.4
P(oranges)=0.1
P(cherries)=0.2
P(plums)=0.3
anyone wanting to play the fruit machine pays 10 cents for a turn. the winning combination are as follow:
orange in 3 windows -1.00
cherries in 3 windows -0.50
orange in 2 windows and cherries in 1 window - 0.80
lemons in 3 windows -0.40

find the expected gain/loss per turn

Answer 1

The expected gain would be
\[\sum_{\large x}x~p(x)\]
where \(x\) is the net amount of money a player would get for a given possible arrangement of the fruits in the windows, and \(p(x)\) is the probability of that event occurring.

The first thing to do would be to find the probability of each listed event occurring. For example, the probability of 3 oranges coming up would be \(P(\text{orange})^3\), or \(0.1^3=.001=p(x)\).

The gain for this particular arrangement (3 oranges) is $1.00. However, the player must pay 10 cents to play, or $0.10, which means the net gain would be \(1.00 - 0.10=0.90=x\).

In the end, the expected value is going to look like this:
\[\text{(net winnings for 3 oranges)}P(\text{3 oranges})\\
+\text{(net winnings for 3 cherries)}P(\text{3 cherries})\\
+\text{(net winnings for 2 oranges, 1 cherry)}P(\text{2 oranges, 1 cherry})\\
+\text{(net winnings for 3 lemons)}P(\text{3 lemons})\\
+\text{(net winnings for any other arrangement)}P(\text{any other arrangement})\]