A fair coin is tossed independently four times. The probability of the event "the number of times heads show up is more than the number of times tails show up" is
A. $$\frac{1}{{16}}$$
B. $$\frac{1}{8}$$
C. $$\frac{1}{4}$$
D. $$\frac{5}{{16}}$$
Answer: Option D
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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