Question

A carnival game gives players a 25% chance of winning every time it has played a player plays the game four times let XP the number of times a player wins in for place what is the most probable value of X what is the probable that the player will win at least once

Answer 1

Answer:

0.6836 = 68.36% probability that the player will win at least once.

Step-by-step explanation:

For each game, there are only two possible outcomes. Either the player wins, or the player loses. The probability of winning a game is independent of any other game, which means that the binomial probability distribution is used 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.

25% chance of winning

This means that $p = 0.25$

Plays the game four times

This means that $n = 4$

What is the probability that the player will win at least once?

This is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = C_{4,0}.(0.25)^{0}.(0.75)^{4} = 0.3164$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.3164 = 0.6836$

0.6836 = 68.36% probability that the player will win at least once.