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:
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:
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:
To calculate the probability of getting 59 or more green balls, we have to calculate the z-value
Then, the probability can be look up with the z-value in a normal distribution table:
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
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:
To calculate the probability of getting 78 or more green balls, we have to calculate the z-value
Then, the probability can be look up with the z-value in a normal distribution table:
There is a probability of P=0.0853 of winning the prize with a 120-balls sample.