Probability of pralines in a box

Question

Milk and Dark chocolate pralines are randomly placed into boxes containing a total of 25 pralines. The colour of each praline is determined by a random mechanism so that on average, 60% of all pralines are dark.

(a) Let the random variable M denote the number of Milk Pralines in a box. What is the distribution of M?

(b) What is the probability that all chocolate pralines in one box are of the same type?

(c) How many boxes need to be selected to have a chance of >90% to have at least one box with at least 15 Milk Pralines?

So far I believe I have managed to understand b). However a) and c) is a bit more difficult for me. What do they mean by asking the distribution?

What i got till now: 25 Pralines total per Box, 60% are Dark Pralines.

       D= Number of Dark pralines/box 

       n=25                

       p=0.6

b) $P[D=25]+P[D=0]={25 \choose 25}p^{25}(1-p)^{0} +{25 \choose 0}p^{0}(1-p)^{25}$

c) $p'=P[M\ge15]= {15 \choose 15}p^{15}(1-p)^{0}= 1-P[M\le14]$ which equals to 0.000470184.

Then $P[X\ge1]=1-P[X=0]= 1-{m \choose 0}p'^{0}(1-p')^{m}= 1-0.999529816^{m} $ Then I'd just use logarithm to find m

$$1-0.999529816^{m}\gt0.9$$ $$0.1\gt0.999529816^{m}$$ $$ln0.1\gt mln0.999529816$$ $$\frac{ln0.1}{ln0.999529816}\lt m$$

The m= 4896.04, SO I'd have to go through 4896 boxes to 
get a box with at least 15 Milk Pralines?

I think my answer is wrong, do I really have to go through 4896 boxes? Have I been using the right methods?

And I'd really appreciate it if someone could explain a) for me.

Answer 1

With regard to (a), you're already using the answer there to solve (b) and (c).

It's simply the binomial distribution (mass probability distribution) of $25$ objects with $p=0.4$:

$$P(M = k) = {25 \choose k}(0.4)^{k}(1-0.4)^{25-k}.$$

Answer 2

a) $M\sim Bin(25,0.4)$

According to my calculator,

b) $p(M=0)+p(M=25)=2.843\times10^{-6}$

c) $p(M\geq15)=1-p(M\leq14)=1-0.9656084819...$

So you should have $$n>\frac{\ln0.1}{\ln0.9656...}=65.79$$ Therefore $n=66$