Q5. X takes the values 0, 1 and 2, with P(X = 1) = 2P(X = 0) and P(X = 2) = 3P(X = 0). Find the distribution and the mean.
If P(X = 0) = k, then k + 2k + 3k = 1, so k = 1/6. The distribution is 1/6, 1/3, 1/2, and μ = 1/3 + 2 × 1/2 = 4/3.
Q6. Two balls are drawn without replacement from a bag with 3 red and 2 black balls. Find the distribution of X, the number of red balls, and its mean.
There are 10 pairs: 1 has no red ball, 3 × 2 = 6 have one, and 3 have two. So P(X = 0) = 1/10, P(X = 1) = 3/5 and P(X = 2) = 3/10, and μ = 3/5 + 3/5 = 6/5.
Q7. Two dice are thrown, and X is the larger of the two numbers. Show that P(X = k) = (2k − 1)/36 for k = 1, 2, …, 6, and find the mean.
X ≤ k when both dice show at most k, which happens in k2 ways, so X = k in k2 − (k − 1)2 = 2k − 1 ways. The counts 1, 3, 5, 7, 9, 11 add up to 36. Multiplying each k by its count gives 1, 6, 15, 28, 45 and 66, which add up to 161, so the mean is 161/36.
Q8. A player pays 1 point to throw a die and wins 4 points on a six, nothing otherwise. Find the expected net gain.
The net gain is 3 with probability 1/6 and −1 with probability 5/6, so its mean is 3/6 − 5/6 = −1/3 point a game.
Q9. Can a random variable have mean 2 and Σ x p = 3?
No. The variance would be 3 − 2 = −1, but Var(X) = Σ (x − μ) p is a sum of terms that are never negative.