Question

Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive.aWhat is the probability that the number will be more than 6 or odd? (Enter your probability as a fraction.)

Answer 1

Start by putting the possible integers your friend can select

$S={}\lbrace1,2,3,4,5,6,7,8,9,10\rbrace$

Then, call the 2 possible events as A and B, and what are the possible integers in each event:

A= Be more than 6

B= The number is odd

$\begin{gathered} A=\lbrace7,8,9,10\rbrace \\ B=\lbrace1,3,5,7,9\rbrace \end{gathered}$

The probability of the union of two events can be calculated as:

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Then,

$\begin{gathered} P(A)=\frac{number\text{ }of\text{ }elements\text{ }in\text{ }A}{number\text{ }of\text{ }elements\text{ }in\text{ }S} \\ P(A)=\frac{4}{10} \\ P(B)=\frac{number\text{ }of\text{ }elements\text{ }in\text{ }B}{number\text{ }of\text{ }elements\text{ }in\text{ }S\text{ }} \\ P(B)=\frac{5}{10} \\ P(A\cap B)=\frac{number\text{ }of\text{ }elements\text{ }in\text{ }A\text{ }and\text{ }B}{number\text{ }of\text{ }elements\text{ }in\text{ }S} \\ P(A\cap B)=\frac{2}{10} \end{gathered}$

Finally,

$\begin{gathered} P(A\cup B)=\frac{4}{10}+\frac{5}{10}-\frac{2}{1^{\prime0}} \\ P(A\cup B)=\frac{7}{10} \end{gathered}$

Answer:

the probability that the number will be more than 6 or odd is: 7/10