- #1

spacetimedude

- 88

- 1

## Homework Statement

Calculate

a) P(exactly three dice have the same number)

b) Calculate the conditional probability P(three of the dice shows six|two of the dice shows 5)

## Homework Equations

## The Attempt at a Solution

a) Say we have 1 as the number we get for three dice. There is 1/6 chance of getting a 1 and 5/6 of getting other numbers. The total sample space is 6^5 for five rolls.

Hence for P(getting three 1's)=[(1/6)^3*(5/6)^2]/[6^5]

But that was just for getting three 1's and there are total of 6 numbers that can be used.

So

P(exactly three dice have the same number)=6*[(1/6)^3*(5/6)^2]/[6^5].

This answer seems to be right, but I would like to know how you can solve this using the binomial coefficent.

My attempt:

3 in 5 rolls will result in a number, which can be expressed as 5C3. Each number has 1/6 chance of being that number. The others have the probability of 5/6. The sample space is still 6^5.

Putting this all together, [6*(5C3)*(1/6)^3*(5/6)^2]/[6^5], which yields a wrong answer.

b) The conditional probability P(three of the dice shows six|two of the dice shows 5) can be written as

P(three of the dice show six|two of the dice show 5)=P(three show 6, two show 5)/P(two show 5).

Why are the events not independent here? Each die is a different entity so can't we just say P(three shows 6, two show 5)=P(three shows 6)*P(two show 5)?

Anyway, P(two shows 5)=[(1/6)^2*(5/6)^3]/[6^5]

I am not sure how to find the probability of the intersection of three show 6 and two show 5. I tried using the propoerty that the numerator of conditional probability can be written as P(three show six)P(two show 5|three show 6) but that got me nowhere.

Any help will be appreciated!