The following is based on flipping a fair coin m times, where m is determined bythe roll of an n-sided die (with sides numbered 1, 2,3 ... n). X = num headsAssume the die is fair and n = 6, that is, a 6-sided die.(a) What is the probability mass function of X?(b) What is the expected value of X?Assume the die is biased and n = 3, with the probability of obtaining a 2 more likely than obtaining either a 1 or 3. That is, the probability of obtaining a two is 2/3 and the probability of obtaining a 1 or 3 equally likely. If X = the number of heads obtained.(c) What is the probability mass function of X?(d) What is the expected value of X?(e) Determine the CDF (Cumulative Distribution Function) of X.

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Asked Oct 8, 2019
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The following is based on flipping a fair coin m times, where m is determined by
the roll of an n-sided die (with sides numbered 1, 2,3 ... n). X = num heads
Assume the die is fair and n = 6, that is, a 6-sided die.
(a) What is the probability mass function of X?
(b) What is the expected value of X?

Assume the die is biased and n = 3, with the probability of obtaining a 2 more likely than obtaining either a 1 or 3. That is, the probability of obtaining a two is 2/3 and the probability of obtaining a 1 or 3 equally likely. If X = the number of heads obtained.
(c) What is the probability mass function of X?
(d) What is the expected value of X?
(e) Determine the CDF (Cumulative Distribution Function) of X.

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Step 1

Note:

Hey there! Thank you for posting the question. However, although the questions are given as parts (a) to (e), parts (a) and (b) form one question, while parts (c), (d) and (e) form a separate question; the only connection between the 2 questions is that the coin is flipped as many times as the number showing on the die. Thus, we have solved the first question [parts (a) and (b)] for you. If you need help with the other question, please re-post it.

Step 2

Part (a):

Here, X denotes the number of heads, when the fair coin is flipped m times.

When a fair, 6-sided die is rolled, there are 6 equally likely possible outcomes. Given that, m is the number obtained on rolling the die; m = 1, 2, …, 6.

For m = 1, the fair coin will be tossed once; the possible values of X will be: X = 0, 1.

For m = 2, the fair coin will be tossed twice; the possible values of X will be: X = 0, 1, 2.

Proceeding in this manner, it can be generalized that:

For the outcome m on the die, the fair coin will be tossed m times; the possible values of X will be: X = 0, 1, …, m.

When a coin is tossed m times, the number of heads (successes), X ~ Binomial (m, p). Now, the probability of success of a fair coin is 1/2.

Thus, for a particular m, (X | m) ~ Bin (m, 1/2); m = 1, 2, …, 6.

The probability mass function (pmf) of (X | m) is:

p (x) = mCx (1/2)m; x = 0, 1, …, m; m = 1, 2, …, 6, and 0 otherwise…….. (1)

Here, the random variable X can be obtained as follows, as the union of mutually exclusive events based on the value of m:

X = (X, m = 1) + (X...

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