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

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Homework Help Overview

The problem involves determining the minimum number of throws of a pair of dice required to achieve a probability greater than 0.95 of rolling a sum of 7 at least once. The subject area pertains to probability theory, specifically dealing with independent events and cumulative probabilities.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the correct interpretation of the probability question, with some initially miscalculating the probability of rolling a 7 on the last throw rather than considering the cumulative probability over multiple throws. There is also a focus on clarifying the distinction between obtaining a 7 on a specific throw versus at least once in a series of throws.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. Some have provided guidance on the correct approach to calculate the probability of rolling at least one 7 in multiple throws, while others are seeking clarification on the implications of the problem statement.

Contextual Notes

Participants are navigating through assumptions about the nature of the probability calculation, particularly the misunderstanding regarding the requirement for the last throw to be a 7. There is an emphasis on the need for clarity in the problem's wording and the implications for the calculations involved.

erisedk
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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.

Homework Equations

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.
 
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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
erisedk said:
But wouldn't the minimum number of throws correspond to a 7 on the last throw?
 
erisedk said:
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 :)
 

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