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)=(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)=(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)=(2)/(10) \end{gathered}`

Finally,

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

Answer:

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