Question

A random variable with geometric distribution is a discrete variable that can attain values of 1, 2, . . . . It is used to model the number of trials of a Bernoulli (binary) test until the first success occurs. So, if a Bernoulli trial can have "success" with probability p and "failure" with probability 1 − p, then the number of iid trials before a success happens follows the geometric distribution. The pmf for this distribution is given by f(k) = (1 − p) k−1p Because if the first success is seen at the kth trial, it means that we have seen k − 1 failures in a row and then saw success in the kth trial. 3 33::487, Spring 2018 Homework 1 Due date: 9/26/18 at 11:50PM 4a) Someone gives you a coin and tells you it may be a fake coin. Furthermore, a fair coin has a 50% probability of heads, while a fake one has only a 20% chance of heads turning up. You toss the coin 3 times in a row, and it turns up tails, and only on the fourth toss, you see the first heads. Based on this observation, would you call this coin fair or fake? Justify your answer. 4b) Suppose we have two scenarios: the probability of success is p1, or probability of success is p2. Given values of 0 ≤ p1 < p2 ≤ 1, what is the minimum number of times failures should occur before the first success in order for you to select scenario 1 over scenario 2?

Answer 1

Answer:

a) You do not have enough to decide the coin is fake

b) You should choose n as the least positive integer such that

$n>\frac{p_1/p_2}{log(\frac{1-p_1}{1-p_2})}+1$

Step-by-step explanation:

a)

If the coin is fair, the probability of head = probability of tail = 0.5 .

If the event of tossing the coin follows the Bernoulli's distribution (also called binomial distribution), then the probability of 1 head in four tosses is

$\binom{4}{1}(0.5)(0.5^3)=4(0.5)^4=0.25=25%$

Since this probability is much less than 50%, it is too early to decide that the coin is not fair.

b)

Suppose you perform the experiment n times in the two scenarios. The probability of having 1 success and n-1 failures in scenario 1 would be

$\binom{n}{1}p_1(1-p_1)^{n-1}$

whereas in scenario 2 would be

$\binom{n}{1}p_2(1-p_2)^{n-1}$

Obviously, you would select scenario 1 if

$\binom{n}{1}p_1(1-p_1)^{n-1}>\binom{n}{1}p_2(1-p_2)^{n-1}$

$p_1(1-p_1)^{n-1}> p_2(1-p_2)^{n-1}$

$\frac{p_1}{p_2}>(\frac{1-p_1}{1-p_2})^{n-1}$

Since the function log(x) is increasing, we can take log on both sides to get

$log(\frac{p_1}{p_2})>(n-1)log(\frac{1-p_1}{1-p_2})$

Since  

$01-p_2\Rightarrow \frac{1-p_2}{1-p_1}$

and

$log( \frac{1-p_2}{1-p_1})$

Therefore  

$n>\frac{p_1/p_2}{log(\frac{1-p_1}{1-p_2})}+1$

So, you should choose n as the least positive integer such that

$n>\frac{p_1/p_2}{log(\frac{1-p_1}{1-p_2})}+1$