Question

The probability of winning on an arcade game is 0.612. If you play the arcade game 21 times, what is the probability of winning no more than 10 times? (Round your answer to 3 decimal places if necessary.)

Answer 1

Answer: 0.111

Step-by-step explanation:

Given: The probability of winning on an arcade game : p=0.612

The number of times  you play the arcade game : n=21

We assume this is a normal distribution , then we have

$\text{Mean}=\mu=np=21(0.612)=12.852$

$\text{Standard deviation}=\sigma=\sqrt{np(1-p)}\\\\=\sqrt{21(0.612)(1-0.612)}\approx2.23$

Let X be a binomial variable .

Then , the z score for x= 10 will be :-

$z=\dfrac{x-\mu}{\sigma}=\dfrac{10-12.852}{2.33}\approx-1.22$

Now, the probability of winning no more than 10 times :_

$P(x\leq10)=P(\leq-1.22)=0.1112324\approx0.111$

Hence, the probability of winning no more than 10 times =0.111