Question

The chance of winning a certain game at a carnival is 2 in 5. If Andy plays the game 12 times, what is the probability that he loses AT MOST 3 times?

Answer 1

1.5267% probability that he loses AT MOST 3 times

Explanation:

For each game that Andy plays, there are only two possible outcomes. Either he wins, or he loses. The probability of winning a game is independent of other games. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

The chance of winning a certain game at a carnival is 2 in 5.

So the chance of losing is (5-2) in 5, that is 3 in 5.

So
`p = (3)/(5) = 0.6`

12 games:

This means that
`n = 12`.

What is the probability that he loses AT MOST 3 times?

`P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)`.

In which:

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(12,0).(0.6)^(0).(0.4)^(12) = 0.000017`

`P(X = 1) = C_(12,1).(0.6)^(1).(0.4)^(11) = 0.000302`

`P(X = 2) = C_(12,2).(0.6)^(2).(0.4)^(10) = 0.002491`

`P(X = 3) = C_(12,3).(0.6)^(3).(0.4)^(9) = 0.012457`

`P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.000017 + 0.000302 + 0.002491 + 0.012457 = 0.015267`

1.5267% probability that he loses AT MOST 3 times