Wiley Edge Probability Quiz-4

The probability of drawing two red balls and two blue balls is (5/10) * (4/9) * (3/8) * (2/7) = 120/5040 = 1/42.

The probability of drawing a red ball and a non-white ball is given by (7/16) * (11/15) + (5/16) * (9/15) + (4/16) * (12/15) = 77/240 + 45/240 + 48/240 = 170/240 = 17/24.

The probability of getting an even number on a fair six-sided die is 3/6 = 1/2, and the probability of getting an odd number is also 1/2. Using the binomial coefficient, the probability of getting exactly two even numbers and four odd numbers is C(6, 2) * (1/2)^2 * (1/2)^4 = 15/64.

The probability of drawing two red balls and one non-red ball is given by
(5/12) * (4/11) * (8/10) = 160/1320 = 4/33.
Similarly, the probability of drawing two blue balls and one non-blue ball is (4/12) * (3/11) * (9/10) = 108/1320 = 9/110. Adding these probabilities together, we get (4/33) + (9/110) = 19/132 + 9/110 = 209/1320 = 19/120.

The probability of drawing at least one red ball and one white ball is equal to 1 minus the probability of drawing all red balls or all white balls.
The probability of drawing all red balls is (6/15) * (5/14) * (4/13) = 120/2730, and the
probability of drawing all white balls is (5/15) * (4/14) * (3/13) = 60/2730.
Hence, the probability of drawing at least one red ball and one white ball is 1 - (120/2730 + 60/2730) = 2550/2730 = 85/91 = 5/6.

The probability of drawing no red balls is given by
(12/16) * (11/15) * (10/14) * (9/13) = 2970/4368 = 135/196.
The probability of drawing exactly one red ball is (7/16) * (9/15) * (8/14) * (7/13) = 1764/4368 = 98/245.

The probability of drawing exactly two red balls is (7/16) * (6/15) * (9/14) * (8/13) = 1512/4368 = 84/261.

Adding these probabilities together, we get (135/196) + (98/245) + (84/261) = 165/196 = 15/18 = 5/6.

The probability of drawing two red balls is (4/9) * (3/8) = 12/72 = 1/6. Similarly, the probability of drawing two blue balls is (3/9) * (2/8) = 6/72 = 1/12. Adding these probabilities together, we get 1/6 + 1/12 = 3/12 = 1/4 = 9/36.

The probability of getting more tails than heads can be calculated by considering the complementary event, which is getting more heads than tails or an equal number of heads and tails. The probability of getting more heads than tails or an equal number of heads and tails is given by (1/2)^8 + C(8, 1) * (1/2)^8 + C(8, 2) * (1/2)^8 = 1/256 + 8/256 + 28/256 = 37/256.
Hence, the probability of getting more tails than heads is 1 - 37/256 = 219/256 = 109/128.

The probability of drawing at least two white balls is equal to 1 minus the probability of drawing no white balls or exactly one white ball. The probability of drawing no white balls is given by (12/16) * (11/15) * (10/14) * (9/13) = 2970/4368 = 135/196. The probability of drawing exactly one white ball is (4/16) * (12/15) * (11/14) * (10/13) = 7920/4368 = 45/49. Hence, the probability of drawing at least two white balls is 1 - (135/196 + 45/49) = 1 - 1260/3388 = 2128/3388 = 532/847 = 133/211 = 133/495.

The probability of drawing one red ball, one white ball, and one blue ball is given by (8/16) * (5/15) * (3/14) + (8/16) * (3/15) * (5/14) + (5/16) * (8/15) * (3/14) + (5/16) * (3/15) * (8/14) + (3/16) * (8/15) * (5/14) + (3/16) * (5/15) * (8/14) = 120/3360 + 120/3360 + 120/3360 + 120/3360 + 120/3360 + 120/3360 = 720/3360 = 12/56 = 3/14.