Answer + Step-by-step explanation:
1) The probability of getting 2 white balls is equal to:

2) the probability of getting 2 white balls is equal to:

3) The probability of getting at least 72 white balls is:

![=\sum^{150}_{k=72} [C^{k}_{150}\times \left( \frac{8}{15} \right)^{k} \times \left( \frac{7}{15} \right)^{150-k}]](https://tex.z-dn.net/?f=%3D%5Csum%5E%7B150%7D_%7Bk%3D72%7D%20%5BC%5E%7Bk%7D_%7B150%7D%5Ctimes%20%20%5Cleft%28%20%5Cfrac%7B8%7D%7B15%7D%20%5Cright%29%5E%7Bk%7D%20%20%5Ctimes%20%5Cleft%28%20%5Cfrac%7B7%7D%7B15%7D%20%5Cright%29%5E%7B150-k%7D%5D)
It should be A. Normal Distribution.
Answer:
$16,666.67
Step-by-step explanation:
PMT= PV*i Where PMT is the withdrawals ,PV is present value and i is the dicounting rate
PMT = $1,000.00
PV= ?
i = 6%
hence $1,000 = PV*6%
PV=1,000/6%
PV = 16,666.67