Slot Machine Probabilities
If there is a slot machine with 4 slots, each with a possibility of being either a cherry, a lemon, or a 7, what is the probability of getting exactly two lemons?
My thinking:
4 slot machine, 3 options: 3*3*3*3 = 81 total options.
getting exactly 2 lemons: 1*1*3*3 = 9
P(LL) = 9/81 = 0.1111
What is wrong with this?
$\endgroup$ 22 Answers
$\begingroup$Your calculation assumes that lemons must be in the first two slots (the $1$s you multiply in) and allows lemons in the last two slots (the $3$s you multiply in), so would accept four lemons and reject 7-7-lemon-lemon. You should compute the number of ways to choose the two slots for the lemons, then require that the other two slots be non-lemons.
$\endgroup$ 8 $\begingroup$As you probably gathered by now from the other answers and comments, there are ${4\choose2}=6$ ways to place the lemons (L's) on the four slots. And then the two remaining slots can either be Cherry or Seven. The table below shows why then you would multiply $6\times4=24$ to get the total arrangements.
The Lemon-only arrangements are... $$ \begin{array}{c|c|c|c|c|c} \text{L-L-x-x} & \text{L-x-L-x} & \text{...}& \text{...}& \text{...} & \text{x-x-L-L} \end{array} $$ So the total arrangements with exactly 2 Lemons would be. $$ \begin{array}{c|c|c|c|c|c} \text{L-L-C-C} & \text{L-C-L-C} & \text{...}& \text{...}& \text{...} & \text{...} \\ \hline \text{L-L-C-7} & \text{...} & \text{...} & \text{...} & \text{...} & \text{...} \\ \hline \text{L-L-7-C} & \text{...} & \text{...} & \text{...} & \text{...} & \text{...} \\ \hline \text{L-L-7-7} & \text{...} & \text{...} & \text{...} & \text{...}& \text{7-7-L-L} \\ \end{array} $$
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