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 RM1.00
cherries in 3 windows RM0.50
orange in 2 windows and cherries in 1 window RM0.80
lemons in 3 windows RM0.40

find the expected gain/loss per turn

Answer 1

Hi, Karl,
Mind defining your terms? That RM you're using has proven to be a distractor. I assume it's a unit of currency other than the dollar.

Answer 2

just use dollar $1.00
$0.50
$0.80
$0.40

Answer 3

I note that the four given probabilities add up to 1.0, as they must.

I then attempted to calculate the probabilities of each of the slot machine outcomes:
for example, the probability of obtaining 3 oranges in a row is 0.1^3, or 0.001; the probability of obtaining two oranges and 1 cherry in a row is (.1^2) * (.2), or 0.002, etc.
Can anyone verify or find fault with this reasoning?

We end up with four probabilities in this manner.

I multiply each of these by the amount of winnings involved:
.001 * $1 (for 3 oranges in a row), and so on,

and then added up the four such products, ending up with an answer (average payout per turn) of $0.03.

I'd welcome comments and constructive criticism of this possible (but untested) solution.