Let X be the number of heads of throwing 6 fair coins once. (a fair coin means p

Question

Let X be the number of heads of throwing 6 fair coins once. (a fair coin means p(head) = p(tail) = 0.5).

(a) What is the probability of P(X = n) for n = 0, 1, 2, .., 6 (7 possible integers)?

(b) Prove that the sum of P(X = n) for n = 0, 1, 2 till 6 is equal to 1.

(c) For which n from 0 to 6, is the probability P (X = n) the maximum? Why?

(d) Let X1 be the number of heads of throwing 6 fair coins the first time, X2 the number of heads of throwing 6 fair coins the second time and W = X1 + X2 be the total number of heads in throwing 6 fair coins twice. Compute P (W = 3) or the probability that the total number of heads is equal to 3 when throwing 6 fair coins two times.

Explanation / Answer

(a)

P(X=0) = 6C0*0.500.56 = 0.015625

P(X=1) = 6C1*0.510.55 = 0.09375

P(X=2) = 6C2*0.520.54 = 0.234375

P(X=3) = 6C3*0.530.53 = 0.3125

P(X=4) = 6C4*0.540.52 = 0.234375

P(X=5) = 6C5*0.550.51 = 0.09375

P(X=6) = 6C6*0.560.50 = 0.015625

(b)

Sum of all probabilities = 0.015625 + 0.09375 + 0.234375 + 0.3125 + 0.015625 + 0.09375 + 0.234375 = 1

(c)

P(X=3) is the maximum because it has the highest numerical value.

(d)

P(W=3) = P(X1 = 0,X2 = 3) + P(X1 = 1,X2 = 2) + P(X1 = 2,X2 = 1) + P(X1 = 3,X2 = 0)

P(X1 = 0,X2 = 3) = P(X1 = 3,X2 = 0) = 0.015625*0.3125 = 0.0048828125

P(X1 = 1,X2 = 2) = P(X1 = 2,X2 = 1) = 0.09375*0.234375 = 0.02197265625

So,

P(W=3) = 2*(0.02197265625+0.0048828125) = 0.0537109375 = 0.0537

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