Question

Let X equal the number of flips of a fair coin that are required to observe

Answer 1

Answer:

(a) I attached a photo with the diagram.

(b) $f(x) = \big( \frac{1}{2} \big)^{x-1}$

(c) 1/4

(d) 4

(e) $k^2/2$

Step-by-step explanation:

(a) I attached a photo with the diagram.

(b) The easiest way to think about this part is in terms of combinatorics. Think about it like this.

To begin with, look at the three each level of the three represents a possible outcome of throwing the coin n-times. If you throw the coin 3 times at the end in total there are 8 possible outcomes. But The favorable outcomes are just 2.

 1  - Your first outcome is HEADS and all the others are different except the last one.

 2  - Your first outcome is TAILS and all the others are different except the last one.

Therefore the probability of the event is

$f(x) = P(X=x) = \frac{2}{2^x} = \frac{1}{2^{x-1}} = \big( \frac{1}{2} \big)^{x-1}$

(c)

P(X = 0) = 0 because it is not possible to have two consecutive tails or heads.

$E(X > 4) = 1 - P(X \leq 3) = 1 - ( P(X = 0 ) + P(X = 1) + P(X = 2) + P(X = 3))\\= 1 - ( \frac{1}{2} + \frac{1}{4} ) = \frac{1}{4}$

(d)

Remember that this is a geometric distribution therefore

$E[X] = p/(1-p)$, in this case  $p = 1/2$ so $E[X] = 1$ and

$E[X+1]^2 = ( E[X] +1 )^2 = (1+1)^2 = 2^2 = 4$

Also

(e)

This is a geometric distribution so its variance is

$Var(X) = \frac{1-p}{p^2} = 1/2 / 1/4 = 1/2$

And using properties of variance

$Var(kX - k ) = Var(kX) = k^2 var(X) = k^2 /2$