# Find the probability of throwing a sum of 7 at least 6 times in 7 throws of a pair of fair dice.?

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• Anonymous
2 months ago

1/6⁷ + (1/6⁶)(5/6) = 1/6⁷ + 5/6⁷ = 6/6⁷ = 1/6⁶ = 1/46656

Explanation if required.

Throwing two fair dice means there are 36 possible outcomes with 6 of them giving a sum of 7.

Probability of getting a sum of 7 is therefore 6/36 = 1/6.

Probability of getting a sum of not-7 is 1 - 1/6 = 5/6.

The probability of throwing a 7 seven times is 1/6⁷.

The probability of of throwing a 7 six times and a non-7 once is (1/6⁶)(5/6) = 5/6⁷

So the probability of getting 7 at least six times in seven throws is just the sum: 1/6⁷ + 5/6⁷ as shown in first line.

(You can do this using the binomial theorem but for simple problems like this it's easiest to go back to the basics IMHO.)

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• In a single throww of two fair 6-sided dice, the probability of getting a sum of 7 is 1/18.

The probability of getting 7 all 7 times is (1/18)^7.

The probability of getting 7 exactly 6 times is 7(17/18) * (1/18)^6

Because you must get something other than 7 in exactly one of the seven throws, and 7 in the remaining 6.

(1/18)^7 + 7(17/18) * (1/18)^6

= 120 / 18^7

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