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Prob(at least one 5 in 120 throws) = 1 - prob(no 5s in 120 throws)= 1 - (5/6)120= 0.999 999 999 69 = almost 1.

Prob(at least one 5 in 120 throws) = 1 - prob(no 5s in 120 throws)= 1 - (5/6)120= 0.999 999 999 69 = almost 1.

Prob(at least one 5 in 120 throws) = 1 - prob(no 5s in 120 throws)= 1 - (5/6)120= 0.999 999 999 69 = almost 1.

Prob(at least one 5 in 120 throws) = 1 - prob(no 5s in 120 throws)= 1 - (5/6)120= 0.999 999 999 69 = almost 1.

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Q: What is the probability of getting a sum of 5 if the dice is tossed 120 times?

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The probability of getting 11 with one throw of 2 dice is 1/6*1/6*2 = 1/18 So the probability of not getting 11 with 1 throw of the dice is 17/18. Tossing the dice 54 times, the probability of not getting 11 54 times is (17/18)54 = 0.0456... So the probability of at least 1 roll of 11 is 1 - 0.0456 = 0.954

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