Probability question considering 6 dice throws

  • Thread starter Uniquebum
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  • #1
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Homework Statement


Dice is thrown 6 times. What's the probability of numbers 5 and 6 showing up at least once.


Homework Equations


This ought to be basic probability calculus but i just can't get my head around this. Some kind of attempt(ish) below. THe answer ought to be 0.418 or 41.8%.


The Attempt at a Solution


Now, i know the amount of different permutations is 6^6. Suppose A = [a number 5 appears] and B = [a number 6 appears].
I think i'd need to do this as
P(A n B) = 1 - P(A^c) - P(B^c) + P(A^c n B^c) where ^c denotes a complement.

Probability of A being false (5 not showing) = (5/6)^6 and it's the same for B. I just can't figure out what P(A^c n B^c) is... It's "something" divided by 6^6 but that's as far as i get... Any help would be appreciated.
 

Answers and Replies

  • #2
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A and B false <=> no 5 and no 6.
Can you calculate the probability of "not 5 and not 6" for a single throw? With that probability, it is easy to calculate "no 5 and no 6" for all 6 throws.
 
  • #3
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Thanks alot! Got it done :). Final calculation being

P(A n B) = 1 - (5/6)^6 - (5/6)^6 + (4/6)^6 = 0.41799...

Anyway, thanks again!
 
  • #4
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Wouldn't the probability be 1 - (4/6)^6 = .9122?

Since the probability of rolling any other number is (4/6) for one roll. And for six rolls it is (4/6)^6.
 
  • #5
haruspex
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Wouldn't the probability be 1 - (4/6)^6 = .9122?

Since the probability of rolling any other number is (4/6) for one roll. And for six rolls it is (4/6)^6.
No, the question aks for the probability of rolling at least one 5 and at least one 6, not at least one 5 or 6.
 

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