Question

Consider the following experiment: Simultaneously toss a fair coin and, independently, roll a fair die.

1. Write out the 12 outcomes that comprise the sample space for this experiment.

2. Let X be a random variable that takes the value of 1 if the coin shows “heads” and the value of 0 if the coin shows “tails”. Let Y be a random variable that takes the value of the “up” face of the tossed die. And let Z = X + Y. Construct the probability mass function for Z.

3. Find the mean for Z.

4. Find the variance for Z.

Answer 1

Let us assume that a fair coin is tossed and a fair die is rolled simultaneously.

In tossing a fair coin we have Head (H) and Tail(T) as the possible outcome.

In rolling a fair die we have 1, 2, 3, 4, 5, 6 as the possible outcome.

a) Let S denotes the sample space which denotes the set of all possible outcomes for tossing a fair coin and rolling a die simultaneously.

Then there are 12 outcomes that comprises the whole sample space given by

S={ (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}.

b) Let X be a random variable that takes the value 0 or 1 if it shows tail and head. Let Y be a random variable which number that comes up in the die.

Let Z=X+Y. We have possible outcome for X and Y as

X={0, 1}

Y={ 1, 2, 3, 4, 5, 6}.

Therefore the possible value of Z is

Z={ 0+1, 0+2, 0+3, 0+4, 0+5, 0+6, 1+1, 1+2, 1+3, 1+4, 1+5, 1+6 }.

Z={ 1, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6, 7}.

Therefore are total number of 12 elements in the set. Therefore the probability mass function is given by ​​​​​ a. The mean of Z is given by

For obtaining the mean of Z, we have to obtain the product z.P(Z=z) given by . Therefore the mean is

b. The variance of Z is given by

Therefore the calculation for variance is given in form of table.

. Therefore the variance based on above two table is given by

. The mean of Z is 4 and Variance of Z is 3.166.