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

Answer:

1.5267% probability that he loses AT MOST 3 times

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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 $p = \frac{3}{5} = 0.6$

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