Chapter 10, Problem 10.7AYK

(a)

To determine

To draw:

The Venn diagram for the given data.

Answer to Problem 10.7AYK

  $0.48$

Explanation of Solution

Given:

	OM	No OM
I	$0.147$	$0.140$
No I	$0.569$	$0.144$

Concept used:

Formula

For A and B

  $P\left(\text{only}A\right)=P\left(A\right)-P\left(A\cap B\right)$

Calculation:

Draw the table

	OM	No OM
I	$0.147$	$0.140$
No I	$0.569$	$0.144$

  $\begin{array}{l}P\left(\text{I}\right)=0.147+0.140=0.287\\ P\left(\text{no}\text{I}\right)=0.569+0.144=0.743\\ P\left(\text{OM}\right)=0.147+0.569=0.716\\ P\left(\text{no}\text{OM}\right)=0.140+0.144=0.284\\ P\left(\text{I}\cap \text{OM}\right)=0.147\\ P\left(\text{no}\text{I}\cap \text{no}\text{OM}\right)=0.144\\ P\left(\text{only}\text{I}\right)=P\left(I\right)-P\left(\text{I}\cap \text{OM}\right)\\ P\left(\text{only}\text{I}\right)=0.287-0.147\\ P\left(\text{only}\text{I}\right)=0.140\\ P\left(\text{only}\text{OM}\right)=P\left(\text{OM}\right)-P\left(\text{I}\cap \text{OM}\right)\\ P\left(\text{only}\text{OM}\right)=0.716-0.147\\ P\left(\text{only}\text{OM}\right)=0.569\end{array}$

Draw the Vann diagram

  

(b)

To determine

To find:

The probability $P\left(\text{OM}\right)$ that a randomly selected American adult diagnosed with diabetes is treated for the given data.

Answer to Problem 10.7AYK

  $P\left(\text{OM}\right)=0.716$

Explanation of Solution

Given:

	OM	No OM
I	$0.147$	$0.140$
No I	$0.569$	$0.144$

Concept used:

Formula

For A and B

  $P\left(\text{only}A\right)=P\left(A\right)-P\left(A\cap B\right)$

Calculation:

Draw the table

	OM	No OM
I	$0.147$	$0.140$
No I	$0.569$	$0.144$

  $\begin{array}{l}P\left(\text{I}\right)=0.147+0.140\\ P\left(\text{I}\right)=0.287\\ P\left(\text{OM}\right)=0.147+0.569\\ P\left(\text{OM}\right)=0.716\end{array}$

(c)

To determine

To find:

The probability that a randomly selected American adult diagnosed with diabetes is treated with either insulin or oral medication for the given data.

Answer to Problem 10.7AYK

  $P\left(\text{I}\text{or}\text{OM}\right)=0.856$

Explanation of Solution

Given:

	OM	No OM
I	$0.147$	$0.140$
No I	$0.569$	$0.144$

Concept used:

Formula

For A and B

  $P\left(\text{only}A\right)=P\left(A\right)-P\left(A\cap B\right)$

Calculation:

Draw the table

	OM	No OM
I	$0.147$	$0.140$
No I	$0.569$	$0.144$

  $\begin{array}{l}P\left(\text{I}\right)=0.147+0.140\\ P\left(\text{I}\right)=0.287\\ P\left(\text{OM}\right)=0.147+0.569\\ P\left(\text{OM}\right)=0.716\\ P\left(\text{I}\text{or}\text{OM}\right)=P\left(\text{I}\cup \text{OM}\right)\\ P\left(\text{I}\text{or}\text{OM}\right)=P\left(\text{I}\right)+P\left(\text{OM}\right)-P\left(\text{I}\cap \text{OM}\right)=0.147\\ P\left(\text{I}\text{or}\text{OM}\right)=0.287+0.716-0.147\\ P\left(\text{I}\text{or}\text{OM}\right)=0.856\end{array}$

(d)

To determine

To find:

The product of $P\left(\text{I}\right)\times P\left(\text{OM}\right)$ and the probability of $P\left(\text{I}\text{and}\text{OM}\right)$ for the given data.

Answer to Problem 10.7AYK

  $P\left(\text{I}\right)\times P\left(\text{OM}\right)=0.205$

Explanation of Solution

Given:

	OM	No OM
I	$0.147$	$0.140$
No I	$0.569$	$0.144$

Concept used:

Formula

For A and B

  $P\left(\text{only}A\right)=P\left(A\right)-P\left(A\cap B\right)$

Calculation:

Draw the table

	OM	No OM
I	$0.147$	$0.140$
No I	$0.569$	$0.144$

  $\begin{array}{l}P\left(\text{I}\right)=0.147+0.140\\ P\left(\text{I}\right)=0.287\\ P\left(\text{OM}\right)=0.147+0.569\\ P\left(\text{OM}\right)=0.716\\ P\left(\text{I}\right)\times P\left(\text{OM}\right)=\left(0.287\right)\times \left(0.716\right)\\ P\left(\text{I}\right)\times P\left(\text{OM}\right)=0.205\end{array}$