Let X1 and X2 be two independent exponentially distributed random variables with means 0.5 and 0.25 respectively. Then Y = min (X1, X2) is
A. exponentially distributed with mean $$\frac{1}{6}$$
B. exponentially distributed with mean 2
C. normally distributed with mean $$\frac{3}{4}$$
D. normally distributed with mean $$\frac{1}{6}$$
Answer: Option A
Related Questions on Probability and Statistics
A coin is tossed 4 times. What is the probability of getting heads exactly 3 times?
A. $$\frac{1}{4}$$
B. $$\frac{3}{8}$$
C. $$\frac{1}{2}$$
D. $$\frac{3}{4}$$
A. 1 and $$\frac{1}{3}$$
B. $$\frac{1}{3}$$ and 1
C. 1 and $$\frac{4}{3}$$
D. $$\frac{1}{3}$$ and $$\frac{4}{3}$$
A. E(XY) = E(X) E(Y)
B. Cov (X, Y) = 0
C. Var (X + Y) = Var (X) + Var (Y)
D. E(X2Y2) = (E(X))2 (E(Y))2
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