Answer:
There is a probability of P₉₀=0.09853 of winning the prize with a 90-balls sample.
There is a probability of P₁₂₀=0.0853 of winning the prize with a 120-balls sample.
Step-by-step explanation:
In this problem, the balls that are removed are then replaced in the total, so the probability of obtaining a green ball is constant and is equal to:
![P_g=\frac{93}{93+65}= 0.5886](https://tex.z-dn.net/?f=P_g%3D%5Cfrac%7B93%7D%7B93%2B65%7D%3D%200.5886)
This type of sampling should be analyzed with the binomial distribution, but given the sample size, it is convenient to use the approximation to a normal distribution.
The parameters of the normal distribution are:
![\mu=P_g*n=0.5886*90=52.9746\\\\\sigma=\sqrt{np(1-p)}=\sqrt{90*0.5886*(1-0.5886)} =4.6684](https://tex.z-dn.net/?f=%5Cmu%3DP_g%2An%3D0.5886%2A90%3D52.9746%5C%5C%5C%5C%5Csigma%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B90%2A0.5886%2A%281-0.5886%29%7D%20%3D4.6684)
To win the prize it is needed at least 65% of green balls in the sample, which means there have to be at least 59 green balls in the sample:
![0.65*90=58.5](https://tex.z-dn.net/?f=0.65%2A90%3D58.5)
To calculate the probability of getting 59 or more green balls, we have to calculate the z-value
![z=\frac{x-\mu}{\sigma}=\frac{59-52.9746}{4.6684}= 1.29](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B59-52.9746%7D%7B4.6684%7D%3D%201.29)
Then, the probability can be look up with the z-value in a normal distribution table:
![P(X \geq 59)=P(z\geq 1.29)=0.09853](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%20%2059%29%3DP%28z%5Cgeq%201.29%29%3D0.09853)
There is a probability of P=0.09853 of winning the prize with a 90-balls sample.
When the sample is of 120 balls, we have to recalculate the parameters.
Normal distribution
![\mu=P_g*n=0.5886*120=70.6320\\\\\sigma=\sqrt{np(1-p)}=\sqrt{120*0.5886*(1-0.5886)} =5.3905](https://tex.z-dn.net/?f=%5Cmu%3DP_g%2An%3D0.5886%2A120%3D70.6320%5C%5C%5C%5C%5Csigma%3D%5Csqrt%7Bnp%281-p%29%7D%3D%5Csqrt%7B120%2A0.5886%2A%281-0.5886%29%7D%20%3D5.3905)
To win the prize it is needed at least 65% of green balls in the sample, which means there have to be at least 78 green balls in the sample:
![0.65*120=78](https://tex.z-dn.net/?f=0.65%2A120%3D78)
To calculate the probability of getting 78 or more green balls, we have to calculate the z-value
![z=\frac{x-\mu}{\sigma}=\frac{78-70.6320}{5.3905}=1.37](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B78-70.6320%7D%7B5.3905%7D%3D1.37%20)
Then, the probability can be look up with the z-value in a normal distribution table:
![P(X \geq 78 )=P(z\geq 1.37)=0.0853](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%2078%20%29%3DP%28z%5Cgeq%201.37%29%3D0.0853)
There is a probability of P=0.0853 of winning the prize with a 120-balls sample.