Question

In Problems 11–16, construct a probability model for each experiment.

Tossing two fair coins, then a fair die.

Short answer

The list of possible outcomes:$S=\{\mathrm{HH}1,\mathrm{HH}2,\mathrm{HH}3,\mathrm{HH}4,\mathrm{HH}5,\mathrm{HH}6,\mathrm{HT}1,\mathrm{HT}2,\mathrm{HT}3,\mathrm{HT}4,\mathrm{HT}5,\mathrm{HT}6,\mathrm{TH}1,\mathrm{TH}2,\mathrm{TH}3,\mathrm{TH}4,\mathrm{TH}5,\mathrm{TH}6,\mathrm{TT}1,\mathrm{TT}2,\mathrm{TT}3,\mathrm{TT}4,\mathrm{TT}5,\mathrm{TT}66\}$

The probability of each outcome will be$\frac{1}{24}$.

Step-by-step solution

Step 1. Given information.

Construct a probability model for experiment: Tossing two fair coins, then a fair die.

Step 2. Construct probability model. 

Considering the following outcomes for two coins and a die.

First coin - A head (H) or a tail (T), Second coin - A head (H) or A tail (T), A die - 1,2,3,4,5 and 6.

The list of possible outcomes in tossing two coins and a die will be:

$S=\{\mathrm{HH}1,\mathrm{HH}2,\mathrm{HH}3,\mathrm{HH}4,\mathrm{HH}5,\mathrm{HH}6,\mathrm{HT}1,\mathrm{HT}2,\mathrm{HT}3,\mathrm{HT}4,\mathrm{HT}5,\mathrm{HT}6,\mathrm{TH}1,\mathrm{TH}2,\mathrm{TH}3,\mathrm{TH}4,\mathrm{TH}5,\mathrm{TH}6,\mathrm{TT}1,\mathrm{TT}2,\mathrm{TT}3,\mathrm{TT}4,\mathrm{TT}5,\mathrm{TT}66\}$

The probability of each outcome will be $\frac{1}{24}$since each of them appeared once in the sample space with 24 total outcomes. If we will add them together, it will be equal to 1.

Step 3. Conclusion.

The list of possible outcomes:

$S=\{\mathrm{HH}1,\mathrm{HH}2,\mathrm{HH}3,\mathrm{HH}4,\mathrm{HH}5,\mathrm{HH}6,\mathrm{HT}1,\mathrm{HT}2,\mathrm{HT}3,\mathrm{HT}4,\mathrm{HT}5,\mathrm{HT}6,\mathrm{TH}1,\mathrm{TH}2,\mathrm{TH}3,\mathrm{TH}4,\mathrm{TH}5,\mathrm{TH}6,\mathrm{TT}1,\mathrm{TT}2,\mathrm{TT}3,\mathrm{TT}4,\mathrm{TT}5,\mathrm{TT}66\}$

The probability of each outcome will be$\frac{1}{24}$.