Question

The probability of winning on an arcade game is 0. 659. if you play the arcade game 30 times, what is the probability of winning exactly 21 times? round your answer to two decimal places

Answer 1

The probability of winning exactly 21 times is:-0.32

Given,

The probability of winning on an arcade game, p = 0.659

Number of times you play arcade game, n = 30

By assuming this as a normal distribution, we get

Mean=μ=30×0.659 =19.77

Standard deviation, σ = $\sqrt{np(1-p)}$

= $\sqrt{30(0.659)(1-0.659)}$

≈ 2.60

Let X be a binomial variable

Then, the z score for x = 21 will be:

z = (x - μ) / σ = $\frac{21-19.77}{2.60}$

≈ 0.47

Now, the probability of winning exactly 21 times

P (x ≥ z) = P (21 ≥ 0.47) = 0.32

Hence the probability of winning exactly 21 times is 0.32

Learn more about probability here:-brainly.com/question/13067945

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