Question

In a certain game of chance, your chances of winning are 0.2. Assume outcomes are independent and that you will play the game five times. What is the probability that you win at most once

Answer 1

Answer:

73.73% probability that you win at most once

Step-by-step explanation:

For each game that you play, there are only two possible outcomes. Either you win, or you lose. The games are independent from each other. 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.

In a certain game of chance, your chances of winning are 0.2.

This means that $p = 0.2$

Assume outcomes are independent and that you will play the game five times.

This means that $n = 5$

What is the probability that you win at most once

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.2)^{0}.(0.8)^{5} = 0.3277$

$P(X = 1) = C_{5,1}.(0.2)^{1}.(0.8)^{4} = 0.4096$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3277 + 0.4096 = 0.7373$

73.73% probability that you win at most once