Probability

Let p = 1/2 be the probability of heads. A wins if they get heads on the 1st, 3rd, 5th... toss. P(A) = p + (1-p)^2 * p + (1-p)^4 * p = p / (1 - (1-p)^2) = (1/2) / (1 - 1/4) = (1/2) / (3/4) = 2/3. B wins the rest, so P(B) = 1 - 2/3 = 1/3.

Total outcomes = 2^6 = 64. Favorable outcomes (at least 4 heads) = 4 heads (6C4=15) + 5 heads (6C5=6) + 6 heads (6C6=1) = 15 + 6 + 1 = 22. Probability = 22/64 = 11/32.

This is a binomial distribution problem with n=5, p=1/2 (odd numbers on a die are 1, 3, 5). P(X=3) = 5C3 * (1/2)^3 * (1/2)^2 = 10 * (1/32) = 10/32 = 5/16.

The probability of the second ball being blue is the same as the first ball if the color of the first is unknown. P(B2) = P(B2|B1)P(B1) + P(B2|R1)P(R1) = (4/8 * 5/9) + (5/8 * 4/9) = 20/72 + 20/72 = 40/72 = 5/9.

Probability = (4/52) * (3/51) * (2/50) * (1/49) = 24 / 6497400 = 1 / 270725.

The probability of picking a red ball is 9 / (9 + p) and a green ball is p / (9 + p). Given 9 / (9 + p) = 3 * (p / (9 + p)), we get 9 = 3p, so p = 3.

There are 4 kings in a standard deck of 52 cards. Probability = 4/52 = 1/13.

Total balls = 14. We choose 5. The number of ways to choose 3 white (from 8) and 2 red (from 6) is C(8,3)*C(6,2) = 56*15 = 840. The ways to choose 4 white and 1 red is C(8,4)*C(6,1) = 70*6 = 420. The ways to choose 5 white is C(8,5) = 56. Total favorable outcomes = 840 + 420 + 56 = 1316. Total possible outcomes = C(14,5) = 2002. Probability = 1316/2002 = 658/1001.

Number cards are the cards numbered 2 through 10, giving 9 ranks in each of the 4 suits. Thus there are 36 number cards out of 52, and the probability is 36/52 = 9/13.

A fair coin has two outcomes, head and tail. The probability of getting a head is 1 favorable outcome out of 2 total outcomes, which is 1/2.