Thread: Probability of rolling a single die and getting a 5 in fewer then 4 rolls.

I am having trouble with this problem......

Find the probability of rolling a single die and getting a 5 in fewer then 4 rolls.
(You stop when you get a 5 )


Thanks in advance

1st roll - 5 (1/6) or not a 5 (5/6)
2nd roll (5/6) - 5 (1/6) or not a 5 (5/6)
3rd roll (5/6)^2 - 5 (1/6) or not a 5 (5/6)
4th roll (5/6)^3 - 5 (1/6) or not a 5 (5/6)
5th roll (5/6)^4 - 5 (1/6) or not a 5 (5/6)

You decide where to stop and what it all means.

So I would stop at the third roll because it is less then 4 and it means that the probability is (5/6)^2?

Good try, but no.

P(5 on 1st roll) = 1/6
p(5 on 2nd roll) = (5/6)*(1/6)
p(5 on 3nd roll) = [(5/6)^2]*(1/6)

You must convince yourself that we have the whole distribution. If we keep rolling, do all the probabilities sum to unity?

so I have to add roll 1, 2 and 3 in order to get the probability of rolling a 5 under 4 rolls so it would be [(5/6)^2]*(1/6)

No, that's just roll 3. You might hit it on either roll 1 or roll 2.

1/6 + (5/6)*(1/6) + [(5/6)^2]*(1/6)

Originally Posted by aprinzen

I am having trouble with this problem......

Find the probability of rolling a single die and getting a 5 in fewer then 4 rolls.
(You stop when you get a 5 )


Thanks in advance

What you're looking for is the probability of getting a number '5' in less than 4 toses.

So that means there's only toss 1, toss2, toss 3.

For toss 1, P(5 on 1st roll) = 1/6
For toss 2, p(5 on 2nd roll) = (5/6)x(1/6)
For toss 3, p(5 on 3nd roll) = [(5/6)^2]x(1/6)

Since it's less than 4 toses, remember that you need to COUNT IN ALL THE FIRST 3 ROLLS.

THEN, you'll get 1/6 + (5/6)x(1/6) + [(5/6)^2]x(1/6) = 91/216

Okay?