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If the probability of winning a slot machine is 5% and you are going to play 500 pulls. Using a normal approximation. What’s the probability that you win less than 40?

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Answer:

0.9986 = 99.86% probability that you win less than 40

Explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:


E(X) = np

The standard deviation of the binomial distribution is:


√(V(X)) = √(np(1-p))

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the zscore of a measure X is given by:


Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
\mu = E(X),
\sigma = √(V(X)).

The probability of winning a slot machine is 5% and you are going to play 500 pulls.

This means that
p = 0.05, n = 500

Mean and Standard deviation:


\mu = E(X) = np = 500*0.05 = 25


\sigma = √(V(X)) = √(np(1-p)) = √(500*0.05*0.95) = 4.8734

What’s the probability that you win less than 40?

Using continuity correction, this is P(X < 40 - 0.5) = P(X < 39.5), which is the pvalue of Z when X = 39.5. So


Z = (X - \mu)/(\sigma)


Z = (39.5 - 25)/(4.8734)


Z = 2.98


Z = 2.98 has a pvalue of 0.9986

0.9986 = 99.86% probability that you win less than 40

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