Probability Distributions

The total number of possible outcomes when two people choose a number from 1 to 25 is 25 * 25 = 625. There are 25 outcomes where they choose the same number (1,1; 2,2; etc.). The probability of winning is 25/625 = 1/25. Therefore, the probability of not winning is 1 - 1/25 = 24/25.

Use the complement: P(at least 1 defective) = 1 - P(no defective) = 1 - P(both good). P(both good) = (16/20) * (15/19) = 240/380 = 12/19. So P(at least 1 defective) = 1 - 12/19 = 7/19.

The phrase AMERICAN PIES has 12 total letters. Count the letters: A appears twice, E appears twice, and I appears twice. So there are 2 E's and 2 I's, giving 4 favorable outcomes out of 12 total letters. This simplifies to 1/3, which is the probability.

This is a classic conditional probability puzzle. For a two-child family, the equally likely possibilities are: BB, BG, GB, GG (where B=boy, G=girl, and order matters). Given that at least one child is a girl, we eliminate BB, leaving three equally likely cases: BG, GB, GG. Only in GG are both children girls. Therefore, P(both girls | at least one girl) = 1/3. Option A is correct. The common mistake is thinking the answer is 1/2, which ignores that BG and GB are distinct cases.

The probability that at least one is alive is 1 minus the probability that both are dead. The probability the husband dies is 0.2 (1-0.8) and the wife dies is 0.1 (1-0.9). 1 - (0.2 * 0.1) = 1 - 0.02 = 0.98.

Assume 100 men and 100 women. Employed men = 90% of 100 = 90. Employed women = 55% of 100 = 55. Total employed = 145. Total people = 200. Probability = 145/200 = 29/40. Other fractions do not match this ratio.

In 'AMERICAN PIES' (12 letters), there are 2 E's and 2 I's, totaling 4 favorable letters. Probability = 4/12 = 1/3. The claimed answer 1/3 is correct.

Probabilities must be between 0 and 1 inclusive. Negative values (like -0.00001) and values greater than 1 (like 1.001) cannot represent probabilities because probability measures the likelihood of an event occurring on a scale from impossible (0) to certain (1).

Probability is calculated as the number of favorable outcomes divided by the total number of outcomes. There are 70 people with type O blood out of a total of 200 people. $70 / 200 = 7 / 20 = 0.35$.

Given that they have at least one daughter, the possible gender combinations are: (Girl, Girl), (Girl, Boy), (Boy, Girl), three equally likely cases. Only one of these three cases has the other child also being a girl. So the probability is 1/3 or approximately 33%. This is a classic conditional probability problem.

This is a binomial probability problem where success = device functions (p=0.9) and we want exactly 7 successes in 10 trials. Use the formula P(X=k) = C(n,k) * p^k * (1-p)^(n-k). C(10,7) = 120, so P = 120 * (0.9)^7 * (0.1)^3 ≈ 0.0574. The key is recognizing that 'functioning properly' probability is 0.9, not 0.1.

Using Bayes' theorem for conflicting predictions: local forecaster says no rain with 2/3 accuracy (so 1/3 chance of rain), federal service says rain with 3/4 accuracy. The weighted probability of rain = (3/4) / (3/4 + 2/3) = (3/4) / (17/12) = 9/17 ≈ 0.53. However, another interpretation gives 13/24 ≈ 0.54. Both round to 3/5 = 0.60 as the closest option.

Given at least one boy, the possible combinations are (B,B), (B,G), (G,B) - 3 equally likely cases. Only (B,B) has two boys, so probability = 1/3. This is counterintuitive - many expect 1/2, but the information 'at least one boy' eliminates (G,G) only.