Q: What is the probability that two dice are thrown and At least 1 die shows exactly 1?

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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.

Related questions

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

It shows the probability that the results of the study are due to mere chance.

1/12

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.