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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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โˆ™ 2013-04-11 09:53:15
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โˆ™ 2013-04-11 09:53:15

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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