11
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36
The probability of rolling a 3 on each roll of an unbiased cuboid die is 1/6 If you mean at least one of the rolls shows a 3 then it is the same as 1 - pr(no roll shows a 3) = 1 - (5/6)⁶⁰ ≈ 1 - 0.0000177 = 0.9999823 If you mean that exactly one 3 is rolled then: Pr(exactly one 3) = 60 × 1/6 × (5/6)⁵⁹ ≈ 0.0002130
This is easiest to solve by working out the probability that no heads show and subtracting this from 1 to give the probability that at least one head shows: Assuming unbiased coins which won't land and stay on their edge, the probability of head = probability of tail = ½ → probability no heads = probability 5 tails = ½^5 = 1/32 → probability of at least one head = 1 - 1/32 = 31/32 = 0.96875 = 96.875 % = 96 7/8 %
useing a punnett square shows two ways to express probability
Out of the 36 combinations, you have to count, or calculate, how many combinations are favorable. In this case: 1-2 2-1 2-2 2-3 3-2 That's 5/36. I am assumining that "one die shows a 2" means "at least one". Otherwise, if you mean "exactly one", you have to exclude the combination 2-2.
The probability is 5/36.
This is the same as 1 minus the probability that neither of them are greater than three. This is 1, minus the probability of getting greater than three, squared. Rolling higher than three has a 1/2 probability, so: P(at least one greater than 3) = 1 - (1/2)2 = 1 - 1/4 = 3/4
The probability of rolling a 6 on each roll of an unbiased cuboid die is 1/6 If you mean at least one of the rolls shows a 6 then it is the same as 1 - pr(no roll shows a 6) = 1 - (5/6)⁶⁰ ≈ 1 - 0.0000177 = 0.9999823 If you mean that exactly one 6 is rolled then: Pr(exactly one 6) = 60 × 1/6 × (5/6)⁵⁹ ≈ 0.0002130
The probability of rolling a 3 on each roll of an unbiased cuboid die is 1/6 If you mean at least one of the rolls shows a 3 then it is the same as 1 - pr(no roll shows a 3) = 1 - (5/6)⁶⁰ ≈ 1 - 0.0000177 = 0.9999823 If you mean that exactly one 3 is rolled then: Pr(exactly one 3) = 60 × 1/6 × (5/6)⁵⁹ ≈ 0.0002130
This is easiest to solve by working out the probability that no heads show and subtracting this from 1 to give the probability that at least one head shows: Assuming unbiased coins which won't land and stay on their edge, the probability of head = probability of tail = ½ → probability no heads = probability 5 tails = ½^5 = 1/32 → probability of at least one head = 1 - 1/32 = 31/32 = 0.96875 = 96.875 % = 96 7/8 %
The probability is 0.5
3
The probability, when the 2-dice total is 5, that one of the two dice shows a two is 1/2. The probability that that die is selected is 1/4.The probability, when the 2-dice total is 5, that one of the two dice shows a two is 1/2. The probability that that die is selected is 1/4.The probability, when the 2-dice total is 5, that one of the two dice shows a two is 1/2. The probability that that die is selected is 1/4.The probability, when the 2-dice total is 5, that one of the two dice shows a two is 1/2. The probability that that die is selected is 1/4.
useing a punnett square shows two ways to express probability
1/12
It shows the probability that the results of the study are due to mere chance.
Out of the 36 combinations, you have to count, or calculate, how many combinations are favorable. In this case: 1-2 2-1 2-2 2-3 3-2 That's 5/36. I am assumining that "one die shows a 2" means "at least one". Otherwise, if you mean "exactly one", you have to exclude the combination 2-2.
A probability density function.