Statistic binomial dist

Question

Can you help me to solve this question pls, I consider that I Will use binomial distrıbutıon but I couldnt

Answer 1

Ad c)

You have to assume, that the prepacked bag contains great many cockies. In this case the probability to pick three black truffles in a row is approximately $0.2\cdot 0.2 \cdot 0.2 =0.2^3=0.008=0.8\text{%}$. This is only a approximation.

Let´s say, that a prepacked bag contains 1000 cockies and 200 of them are black truffles. The probability to pick three black truffles in a row is then $\frac{200}{1000}\cdot \frac{199}{999} \cdot \frac{198}{998}=0.007904...=0.7904...\text{%}$

The more cockies are in the bag, the better is the approximation. The approximation has to be done, because we don´t know how many cockies are in one bag.

Answer 2

a) The probability is 0.4. I hope there is no need to discuss it anymore.

b) Again very simple. The answer is 0.4 + 0.2 = 0.6.

c) You have to realize what is the probability of three independent events. According to the definition of independence, it is product of the events. In your case it is 0.2*0.2*0.2 = 0.008.