S1 Probability - Binomial & Geometric Distribution

[AntSC]

Homework Statement

Four players play a board game which requires them to take it in turns to throw two fair dice. Each player throws the two dice once in each round. When a double is thrown the player moves forward six squares. Otherwise the player moves forward one square

Homework Equations

Find:
a) the probability that the first double occurs on the third throw of the game
b) the probability that exactly one of the four players obtains a double in the first round
c) the probability that a double occurs exactly once in 4 of the first 5 rounds

The Attempt at a Solution

a) A geometric problem.
$\mathbb{P}\left ( X=3 \right )=\frac{1}{6}\left ( \frac{5}{6} \right )^{2}=0.1157$

b) A binomial problem.
$\mathbb{P}\left ( X=1 \right )=\binom{4}{1}\frac{1}{6}\left ( \frac{5}{6} \right )^{3}=0.3858$

I've checked the answers for part a) and b) and they're correct.

c) This part i have no idea. I've wracked my brains for ages and i can't get clear about making sense of 'once in 4
of the first 5 rounds'. Any help would be brilliant

 

[haruspex]

I think it means that in four of the first five rounds a double happens exactly once. Could be any four of the five. Not sure what should be assumed for the other round... any result at all? No doubles? Any number of doubles except one?

 

[AntSC]

Isn't saying 'once in four of the first five', the same as saying 'once in the first five'?

 

[haruspex]

Isn't saying 'once in four of the first five', the same as saying 'once in the first five'?

Not if it means "once in each of some four of the first five".

 

[AntSC]

So i take the probability of once in four and then multiply it by how many ways i can get 4 rounds from a total of 5 rounds, namely $\binom{5}{4}$ ?

 

[AntSC]

Or $\binom{5}{4}\mathbb{P}\left ( X=1 \right )$ ?
This is clearly wrong. But it's the closest to making sense to me right now.

 

[haruspex]

So i take the probability of once in four and then multiply it by how many ways i can get 4 rounds from a total of 5 rounds, namely $\binom{5}{4}$ ?

No, you need it to happen in exactly four of the five rounds.

 

[AntSC]

I get it now.
It's a new binomial problem with parameters $X\sim B\left ( 5, 0.3858 \right )$
So $\mathbb{P}\left ( X=4 \right )=\binom{5}{4}p^{4}q=0.068$
Checked the answer, and it agrees.
Thanks for the help