Question

A coin with probability of heads equal to .6 is tossed a first series of 10 tosses.

Let X be the number of heads and let Y be the number of tails obtained.

(a) (1 POINT) Argue with a short sentence that the covariance Cov(X, Y) should be negative.
(b) (1 POINT) Find Cov(X,Y).

In a second series of tosses, the same coin is tossed as many times as the number of heads in the first series.

(c) (3 POINTS) Find the expected number of heads in the first and second series of tosses.

(d) (3 POINTS) Find the probability that the number of heads in the second series is 0.

Answer 1

(a)

Here, we are given ten tosses and X and Y denotes the no. of heads and tails obtained respectively. Clearly, if we obtain high no. of heads (X), then we obtain low number of tails (Y) and vice-versa. Thus, Cov(X,Y) is negative. [ANSWER]

Explanation:

Covariance is a measure of how two variables vary jointly. If high values one variable correspond to high values of another variable and similar behavior for low values then covariance is positive. If high values of one variable correspond to low variables of the other variable and vice-versa then covariance is negative.

(b)

Clearly, (No. of heads obtained in 10 tosses) + (No. of tails obtained in 10 tosses) = 10

=> X + Y = 10 ..........(1)

Now, consider:

Now, since X is the no. of heads obtained in 10 tosses of a coin with probability of heads on each toss equal to 0.6. Thus,

X ~ Binomial(n = 10, p = 0.6) and variance of X is given by:

Var(X) = n*p*(1-p) = 10*0.6*0.4 = 2.4 ............. (3)

From equations (2) and (3), we get:

Cov(X,Y) = -2.4 [ANSWER]

(c)

We know that: X ~ Binomial(n = 10, p = 0.6)

Thus, the expected no. of heads in the first series:

E(X) = n*p = 10*0.6 = 6 [ANSWER]

Let Z denote the no. of heads obtained in the second series of tosses. Now, in the second series of tosses, the coin is tossed X times and the probability of head in each toss is 0.6.

Thus, Z|X ~ Binomial(m = X, q = 0.6)

Now, the expected no. of heads in the second series of tosses:

(d)

The distribution of Z|X is given by:
Z|X ~ Binomial(m = X, q = 0.6)

Thus, the PMF of Z|X is given by:

and the distribution of X is given by:

X ~ Binomial(n = 10, p = 0.6)

Thus, the pmf of X is given by:

Now, the probability that the number of heads in the second series is 0 is given by:

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