You have a coin with unknown probability p of coming up heads. You wish to generate a random variable which takes the values 0 and 1, each with probability 1/2. Assume 0 < p < 1. You adopt the following procedure. You start by flipping the coin twice. If both flips produce the same side of the coin, you start again. If the result of the first flip is different from the result of the second flip, you report the result of the first flip, and you are finished (this is a trick originally due to John von Neumann). (a) Show that, in this case, the probability of reporting heads is 1/2. (b) What is the expected number of flips you must make before you report a result?

(a) Hence, the probability of reporting heads is 1/2.

(b) Therefore, the expected number of such trial is 1 / 2p (1 –p).

Step-by-step explanation:

Solution:

To generate a random variable takes value 0 and 1.

Let 0 < p <1,

A coin is flipped twice.

(a)Probability of reporting head is 1/2.

The result of two toss is either HT or TH.

In this case we used conditional property, Given that you got either HT or TH,

Probability of occurring head or tail:

P (1ST = H| HT or TH]

= P (H) / P (HT or TH)

= P (HT) / P (HT or TH)

= P (1-P) / P (1-P) + (1-P)P

= P (1-P) / 2P (1-P)

=1/2

Hence, the probability of reporting heads is 1/2.

(b)Expected number of flip before report a result.

The probability of reporting a result on each trial is:

P (HT or TH)

=P (HT) + P (TH)

=P (1-P) + (1 –P) P

= 2P (1 –P)

Since, E(X) = 1 / P

Therefore, the expected number of such trial is 1 / 2p (1 –p).

answer: c

(2, -4)

goes 7 to the

(9, 4)

easy