Similar to the above question. Now suppose there are two coins looks exactly the
ID: 3357076 • Letter: S
Question
Similar to the above question. Now suppose there are two coins looks exactly the same. One a fair coin, and the other one has a probability of 0.6 landing a head. I choose one of them with equal probability, and throw it four times, and get paid a dollar every time I get a head. Let X be my payoff after first throw, and Y be my total payoff. Calculate the following:
(a) Joint probability distribution function of X and Y ;
(b) Marginal probability distribution function of X and Y ;
(c) Conditional probability distribution functionof Y conditioned on X = 1
Explanation / Answer
The Payoff on 1st throw can be $0 or $1 based on head comes on first throw or not.
The probability of head coming on 1st throw = P(Choose fair coin) * 0.5 + P(Choose unfair coin) * 0.6
= 0.5 * 0.5 + 0.5 * 0.6 = 0.55
The probability of tail coming on 1st throw = 1 - 0.55 = 0.45
So, PMF of X is
0 with p = 0.45
1 with p = 0.55
Now, we know that the probability of head in any trial is 0.55
So, by binomial distribution formula, the PMF of Y for X = 0 (when 1st throw is tail) can be written as,
0 (No heads of 3 trials) p = 3C0 * 0.550 * (1 - 0.55)3-0 = 0.45 * 0.091125 = 0.04100625
1 (1 heads of 3 trials) p = 3C1 * 0.551 * (1 - 0.55)3-1 = 0.45 * 0.334125 = 0.1503562
2 (2 heads of 3 trials) p = 3C2 * 0.552 * (1 - 0.55)3-2 = 0.45 * 0.408375 = 0.1837688
3 (3 heads of 3 trials) p = 3C3 * 0.553 * (1 - 0.55)3-3 = 0.45 * 0.166375 = 0.07486875
4 p = 0
The PMF of Y for X = 1 (when 1st throw is head) can be written as,
0 p = 0
1 (No heads of 3 trials) p = 3C0 * 0.550 * (1 - 0.55)3-0 = 0.55 * 0.091125 = 0.05011875
2 (1 heads of 3 trials) p = 3C1 * 0.551 * (1 - 0.55)3-1 = 0.55 * 0.334125 = 0.1837688
3 (2 heads of 3 trials) p = 3C2 * 0.552 * (1 - 0.55)3-2 = 0.55 * 0.408375 = 0.2246063
4 (3 heads of 3 trials) p = 3C3 * 0.553 * (1 - 0.55)3-3 = 0.55 * 0.166375 = 0.09150625
Joint probability distribution function of X and Y is,
b)
Marginal probability distribution function of X and Y is given in above table,
P(X = 0) = 0.45
P(X = 1) = 0.55
Marginal distribution function of Y is,
Y = 0 with p = 0.04100625
Y = 1 with p = 0.2004749
Y = 2 with p = 0.3675376
Y = 3 with p = 0.2994751
Y = 4 with p = 0.09150625
c)
Conditional probability distribution functionof Y conditioned on X = 1
Given X = 1,
The PMF of Y for X = 1 (when 1st throw is head) can be written as,
Y = 0 p = 0
Y = 1 p = 0.05011875
Y = 2 p = 0.1837688
Y = 3 p = 0.2246063
Y = 4 p = 0.09150625
X Y 0 1 2 3 4 Total 0 0.04100625 0.1503562 0.1837688 0.07486875 0 0.45 1 0 0.05011875 0.1837688 0.2246063 0.09150625 0.55 Total 0.04100625 0.2004749 0.3675376 0.2994751 0.09150625 1Related Questions
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