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Expert-verified Found in: Page 865 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # In Problems 11–16, construct a probability model for each experiment.Tossing a fair coin, a fair die, and then a fair coin.

The list of possible outcomes:

$S=\left\{\mathrm{H}1\mathrm{H},\mathrm{H}2\mathrm{H},\mathrm{H}3\mathrm{H},\mathrm{H}4\mathrm{H},\mathrm{H}5\mathrm{H},\mathrm{H}6\mathrm{H},\mathrm{H}1\mathrm{T},\mathrm{H}2\mathrm{T},\mathrm{H}3\mathrm{T},\mathrm{H}4\mathrm{T},\mathrm{H}5\mathrm{T},\mathrm{H}6\mathrm{T},\mathrm{T}1\mathrm{H},\mathrm{T}2\mathrm{H},\mathrm{T}3\mathrm{H},\mathrm{T}4\mathrm{H},\mathrm{T}5\mathrm{H},\mathrm{T}6\mathrm{H},\mathrm{T}1\mathrm{T},\mathrm{T}2\mathrm{T},\mathrm{T}3\mathrm{T},\mathrm{T}4\mathrm{T},\mathrm{T}5\mathrm{T},\mathrm{T}6\mathrm{T}\right\}$

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

See the step by step solution

## Step 1. Given information.

Construct a probability model for experiment: Tossing a fair coin, a fair die, and then a fair coin.

## Step 2. Construct probability model.

Considering the following outcomes for a coins, a die and another coin:

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

$S=\left\{\mathrm{H}1\mathrm{H},\mathrm{H}2\mathrm{H},\mathrm{H}3\mathrm{H},\mathrm{H}4\mathrm{H},\mathrm{H}5\mathrm{H},\mathrm{H}6\mathrm{H},\mathrm{H}1\mathrm{T},\mathrm{H}2\mathrm{T},\mathrm{H}3\mathrm{T},\mathrm{H}4\mathrm{T},\mathrm{H}5\mathrm{T},\mathrm{H}6\mathrm{T},\mathrm{T}1\mathrm{H},\mathrm{T}2\mathrm{H},\mathrm{T}3\mathrm{H},\mathrm{T}4\mathrm{H},\mathrm{T}5\mathrm{H},\mathrm{T}6\mathrm{H},\mathrm{T}1\mathrm{T},\mathrm{T}2\mathrm{T},\mathrm{T}3\mathrm{T},\mathrm{T}4\mathrm{T},\mathrm{T}5\mathrm{T},\mathrm{T}6\mathrm{T}\right\}$

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:

$S=\left\{\mathrm{H}1\mathrm{H},\mathrm{H}2\mathrm{H},\mathrm{H}3\mathrm{H},\mathrm{H}4\mathrm{H},\mathrm{H}5\mathrm{H},\mathrm{H}6\mathrm{H},\mathrm{H}1\mathrm{T},\mathrm{H}2\mathrm{T},\mathrm{H}3\mathrm{T},\mathrm{H}4\mathrm{T},\mathrm{H}5\mathrm{T},\mathrm{H}6\mathrm{T},\mathrm{T}1\mathrm{H},\mathrm{T}2\mathrm{H},\mathrm{T}3\mathrm{H},\mathrm{T}4\mathrm{H},\mathrm{T}5\mathrm{H},\mathrm{T}6\mathrm{H},\mathrm{T}1\mathrm{T},\mathrm{T}2\mathrm{T},\mathrm{T}3\mathrm{T},\mathrm{T}4\mathrm{T},\mathrm{T}5\mathrm{T},\mathrm{T}6\mathrm{T}\right\}$

The probability of each outcome will be $\frac{1}{24}$. ### Want to see more solutions like these? 