Question

A die is thrown 8 times. What is the probability that 3 will show i) exactly 2 times ii) at least 2 times ii) at most once​

Answer 1

The minimum is when p=0 or p=5… If the die never gets a 5, then it will never get a 5 in the fourth throw, and less being it for the second time. If the die always gets a 5 (p=1), then in the fourth throw it will always get 5, but it will be for the fourth time, not the second time.

What would be the p to get a maximum probability?

Derive it:

(1-p)^2 = p^2 -2p + 1

p^2 * (1-p)^2 = p^4 -2p^3 +p^2

derivative:

4*p^3 - 6*p^2 +2p= 2p*(2p^2 - 3p + 1)

it is 0 when p = 0 or

p = ( 3 +/- sqrt(9 - 8))/4 = 2/4 or 1

p = 0 and p=1 are minima

p = 1/2 is the maximum.

In this case the probability is 3/16 = 18.75

Answer 2

Answer:

Here's the solution mate :

Step-by-step explanation:

Refer to the image pls.