Probability

Drawing with replacement means the probability of drawing an ace is 4/52 = 1/13 for each draw. The probability of two aces is (1/13) * (1/13).

The multiples of 2 or 3 from 1 through 11 are 2, 3, 4, 6, 8, 9, and 10, giving seven favorable outcomes. The probability is therefore 7/11.

Outcomes summing to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), giving 6 favorable outcomes out of 36. The probability is 6/36 = 1/6.

Given that exactly 3 of the 4 positions are white, symmetry makes each position equally likely to be one of the white positions. Thus, the probability that the third draw is white is 3 out of 4. This remains true for both sampling with replacement and sampling without replacement under the stated condition.

Possible outcomes for two coins: {HH, HT, TH, TT}. Total = 4. Favorable (HH) = 1. Probability = 1/4.

There are 13 spades and 12 face cards (excluding the 3 spade face cards already counted). Total favorable outcomes = 13 + 9 = 22. Probability = 22/52 = 11/26.

For binomial probability P(X=k) = C(n, k) * p^k * (1-p)^(n-k). Since p=0.5, P(X=6) = P(X=8) implies C(n, 6) = C(n, 8). Thus 6 + 8 = n, so n = 14.

Probability of Head = 1/2. Probability of odd number on a 6-faced die (1, 3, 5) = 3/6 = 1/2. Since these are independent events, the combined probability is 1/2 * 1/2 = 1/4.

Odd numbers on a die are {1, 3, 5}. Prime numbers on a die are {2, 3, 5}. Prime numbers greater than 3 are {5}. Given that an odd number came up, the sample space is {1, 3, 5}. Only {5} is a prime number greater than 3. Thus, the probability is 1/3.

The total outcomes for two dice are 36. Sums that are prime are 2, 3, 5, 7, 11. Counting the pairs for these sums: (1,1), (1,2), (2,1), (1,4), (4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3), (5,6), (6,5). There are 15 prime sums, so 21 sums are not prime. The probability is 21/36 = 7/12.

Total outcomes for two dice = 36. Even sums: (1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6) = 18 outcomes. Sums less than 5: (1,1), (1,2), (1,3), (2,1), (2,2), (3,1) = 6 outcomes. The union includes (1,1), (1,3), (2,2), (3,1) which are in both sets. Total unique outcomes = 18 + 6 - 4 = 20. Probability = 20/36 = 5/9.

Total outcomes for 3 dice is 6^3 = 216. Outcomes summing to 5 are (1,1,3), (1,3,1), (3,1,1), (1,2,2), (2,1,2), (2,2,1). There are 6 such outcomes. Probability = 6/216 = 1/36.

Probability of getting at least one head in n tosses is 1 - (1/2)^n. We need 1 - (1/2)^n >= 0.9, which means (1/2)^n <= 0.1. For n=3, 1/8 = 0.125. For n=4, 1/16 = 0.0625. Thus, 4 tosses are required.