Hi folks,

This probability question has me puzzled!

In order to start a game of chance a player throws a dice until he obtains a 6. He then records whatever score he obtains on subsequent throws. For example: Throws of 2,4,3,6,4,6,2,5 give recorded scores of 4,6,2,5.

Calculate the probability that

a) the first score recorded is that of the player's fourth throw

b) the player does not record a score in his first 5 throws

The player has 7 throws. Calculate the probability that he will have recorded

c) exactly three 5s and a 3

d) a total score of 3

So:The first bit is ok. If the first score recorded is on the 4th throw, then the 3rd throw must have been a six. For this to happen the first throw and second throw must not be six, so

P(6 on 3rd throw) = 5/6. 5/6. 1/6 = 25/216

The second bit is where I get stuck. The problem is with the wording. Does this mean that I am to assume a six is thrown on the 5th try, so that scores can be recorded from the 6th throw onwards or do I just assume that the first 5 throws do not produce a six?

i.e. P(no 6 in 5 throws) = 5/6.5/6.5/6.5/6.1/6.

or = 5/6.5/6.5/6.5/6.5/6

in fact neither give the correct answer! Any suggestions here?

For part c we have to calculate the probability of throwing:

x.x.6.3.5.5.5. where x is between 1 and 5

x.x.6.5.3.5.5.

x.x.6.5.5.3.5

x.x.6.5.5.5.3

P(score of 3) = 5/6.5/6.1/6.1/6.1/6.1/6.1/6.

since selecting any number on a dice is a 1/6. Then multiply by 4 to cover all the permutations. This gives the correct answer but now I am stuck on d!

To score 3, I reckon we have the following possibilities:

x.x.x.x.x.6.3 i.e. 5/6.5/6.5/6.5/6.5/6.1/6.1/6.

x.x.x.x.6.2.1 ditto

x.x.x.x.6.1.2 ditto

x.x.x.6.1.1.1 ditto.

Then add up the probabilities.This gives the wrong answer and makes me doubt whether my approach in part c is correct.

Any help with this would be appreciated!

best regards