Question

The probability of winning something on a single play at a slot machine is 0.11. After 4 plays on the slot machine, what is the probability of winning at least once

Answer 1

Step-by-step explanation:

The probability of winning at least once is equal to 1 minus the probability of not winning any.

P(x≥1) = 1 - P(x=0)

P(x≥1) = 1 - (1-0.11)^4

P(x≥1) = 1 - (0.89)^4

P(x≥1) = 0.373

The probability is approximately 0.373.

Answer 2

Answer:

37.26% probability of winning at least once

Step-by-step explanation:

For each play, there are only two possible outcomes. Either you win, or you do not win. The probability of winning on eah play is independent of other plays. 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 probability of winning something on a single play at a slot machine is 0.11.

This means that $p = 0.11$

After 4 plays on the slot machine, what is the probability of winning at least once

Either you do not win any time, or you win at least once. The sum of the probabilities of these events is decimal 1. So

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

We want $P(X \geq 1)$. So

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

In which

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

$P(X = 4) = C_{4,0}.(0.11)^{0}.(0.89)^{4} = 0.6274$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6274 = 0.3726$

37.26% probability of winning at least once