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

  • Thread starter erisedk
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  • #1
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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.
 

Answers and Replies

  • #2
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.
 
  • #3
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But wouldn't the minimum number of throws correspond to a 7 on the last throw?
 
  • #4
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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?
 
  • #5
Orodruin
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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).
 
  • #6
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Okey got it! Thanks :)
 

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