Probability

Total cards = 52. Spade = 13. Ace = 4. One ace is a spade, so total favorable outcomes = 13 + 3 = 16. Probability of winning = 16/52 = 4/13. Probability of losing = 1 - 4/13 = 9/13. Odds against winning = (9/13) / (4/13) = 9:4.

The probability that neither throw shows 4 is (5/6)^2 = 25/36. Therefore, the probability of at least one 4 is 1 - 25/36 = 11/36.

Let total members be 1. Women = 1/3, so Men = 2/3. Indians = 2/3, so Non-Indians = 1/3. Non-Indians who are women = 3/8 of 1/3 = 1/8. Non-Indians who are men = 1/3 - 1/8 = 5/24. The probability a man is a non-Indian is (5/24) / (2/3) = 5/16 = 0.3125.

Probability of no 6 in one throw = 5/6. Probability of no 6 in 3 throws = (5/6)^3 = 125/216. Probability of at least one 6 = 1 - 125/216 = 91/216.

Total balls = 2+5+1+7 = 15. Non-blue balls = 2+1+7 = 10. Probability of picking 2 non-blue balls = (10/15) * (9/14) = (2/3) * (9/14) = 18/42 = 3/7.

Total balls = 12. Ways to pick 2 balls = 12C2 = 66. Ways to pick 2 red = 5C2 = 10. Ways to pick 2 black = 7C2 = 21. Total favorable = 10 + 21 = 31. Probability = 31/66.

Total balls = 3 + 4 + 5 = 12. Red or green balls = 3 + 5 = 8. Probability = 8 / 12 = 2 / 3.

Total balls = 15. Ways to draw 4 balls = 15C4 = 1365. Ways to draw 0 green balls = 11C4 = 330. Probability of 0 green = 330/1365 = 22/91. Probability of at least one green = 1 - 22/91 = 69/91. Since 69/91 is not listed, the correct answer is 'None of these'.

The probability of rolling 1 on the first roll is 1/6, and the probability of rolling 4 on the second roll is also 1/6. Since the rolls are independent, multiply the probabilities: 1/6 × 1/6 = 1/36.

There are 12 face cards in a deck of 52. The probability of drawing two face cards is (12/52) * (11/51) = (3/13) * (11/51) = 33 / 663 = 11 / 221.

The probability of selecting three men is C(5,3)/C(9,3) = 10/84 = 5/42. Therefore, the probability of including at least one woman is 1 - 5/42 = 37/42. Selecting a head does not change the committee composition probability.

Composite numbers on a die: 4, 6 (2 outcomes). Prime numbers: 2, 3, 5 (3 outcomes). Probability = (2/6) * (3/6) = (1/3) * (1/2) = 1/6.

The probability of getting at least one head in n tosses is 1 - (1/2)^n. We need 1 - (1/2)^n > 0.95, which means (1/2)^n < 0.05. For n=4, (1/2)^4 = 0.0625. For n=5, (1/2)^5 = 0.03125. Thus, 5 tosses are required.