Question

An experiment is performed with a coin which has a head on one side and a tail on the other side. The coin is flipped repeatedly until either exactly two heads have appeared or until the coin has been flipped a total of six times, whichever occurs first. Let X denote the number of times the coin is flipped. The probability that the coin comes up heads on any given flip is denoted as p. For parts (a) to (e), assume that p= 0.7. (a) [2 marks] Find the probability that the coin was flipped six times, i.e., find P(X = 6) (b) [2 marks] Given that the coin was flipped six times, find the probability that two heads appeared in the six flips. (c) [4 marks] Determine the probability distribution of X. Do not round your answers when reporting the probabilities. (d) [1 mark] Calculate the expected value of the number of times the coin was flipped, i.e., calculate E(X). (e) [2 marks] Calculate the variance of the number of times the coin was flipped, i.e., calculate V(X).

Answer 1

(a)

Let Y be the number of heads on X number of flips.

Y | X ~ Binomial(n = X, p = 0.7)

Probability that the coin was flipped six times = Probability of no heads when the coin was flipped six times + Probability of exactly one head when the coin was flipped six times + Probability of exactly one head when the coin was flipped five times * Probability of head when the coin was flipped 6th time

(3) + 0.7" + 0.35 + (3) + 0.71 +0.35 + (3)+c * 0.71 +0.3 +0.7

10.35 +6+0.7 +0.35+5+0.72 +0.34

= 0.03078

(b)

Given that coin was flipped six times, probability that two heads appeared in the six flips

= Probability of exactly one head when the coin was flipped five times * Probability of head when the coin was flipped 6th time / Probability that the coin was flipped six times

(5*0.72 * 0.3)/0.03078

= 0.6447368

(c)

The support of X is 2, 3, 4, 5, 6 (We need at least 2 flips wo get two heads)

The probability distribution of X is,

2 P(X = 2) = P(Y = 2X = 2) = *0.72 * 0.30 = 0.49 2

P(X = 3) = Probability of exactly one head when the coin was flipped two times * Probability of head when the coin was flipped 3rd time

$P(X = 3) = P(Y = 1 | X = 2) * 0.7 = \binom{2}{1} * 0.7^1 * 0.3^1 * 0.7 = 0.294$

P(X = 4) = Probability of exactly one head when the coin was flipped three times * Probability of head when the coin was flipped 4th time

$P(X = 4) = P(Y = 1 | X = 3) * 0.7 = \binom{3}{1} * 0.7^1 * 0.3^2 * 0.7 = 0.1323$

P(X = 5) = Probability of exactly one head when the coin was flipped four times * Probability of head when the coin was flipped 5th time

$P(X = 5) = P(Y = 1 | X = 4) * 0.7 = \binom{4}{1} * 0.7^1 * 0.3^3 * 0.7 = 0.05292$

The probability dostribution of X is,

X	2	3	4	5	6
P(X)	0.49	0.294	0.1323	0.05292	0.03078

(d)

E(X) = 2 * 0.49 + 3 * 0.294 + 4 * 0.1323 + 5 * 0.05292 + 6 * 0.03078 = 2.84048

(e)