# Probability problem -- Number of throws of a pair of dice to get a 7

## Homework Statement

The minimum number of throws of a pair of dice so that the probability of getting the sum of the digits on the dice equal to 7 on atleast one throw is greater than 0.95, is n. Find n.

## The Attempt at a Solution

There are 6 possible ways of getting a sum of 7.
So, P(7) = 6/36 = 1/6 and P(not 7) = 5/6
For minimum number of throws, on the last throw the sum should be 7.
So,
(5/6)n-1 (1/6) > 0.95
(5/6)n-1 > 5.7

Solving for n gives an incorrect answer.

## Answers and Replies

Orodruin
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You are computing the probability of rolling 7 on the nth roll. Not the probability of obtaining at least one seven in n rolls.

But wouldn't the minimum number of throws correspond to a 7 on the last throw?

Doing what you said does give the right answer--
1- (5/6)n < 0.95

n= 17

But I don't understand
But wouldn't the minimum number of throws correspond to a 7 on the last throw?

Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
But wouldn't the minimum number of throws correspond to a 7 on the last throw?

They are not asking for having a 7 on the last throw. They are asking how many throws you have to make to have a 95% chance of getting at least one seven, the seven can be in any of those throws (or several).

Okey got it! Thanks :)