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.
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
1 out of 6 times,or 16.67% probability.
It is 0.9034, approx.
There are 6 outcomes, a 2 is one of them so the probability is 1/6.
1 in 36
If they are fair dice, the probability is 0.0032If they are fair dice, the probability is 0.0032If they are fair dice, the probability is 0.0032If they are fair dice, the probability is 0.0032
5/36 or 13.9%.
These are independent one has no bearing on the other
The probability that the sum is seven all three times is 1/216.
The probability is 90/216 = 5/12
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
1 out of 6 times,or 16.67% probability.
It is 0.9034, approx.
There are 6 outcomes, a 2 is one of them so the probability is 1/6.
If they are normal dice, the probability is 0.
1 in 36
1/6