Question

Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive.

What 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

`\text{S}=\{1,2,3,4,5,6,7,8,9,10\}`

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

`\text{A}=\{7,8,9,10\}`

`\text{B}=\{1,3,5,7,9\}`

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

`\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-P(\text{A}\cap\text{B})`

Then,

`\text{P(A)}=\frac{\text{number of elements in A}}{\text{number of elements in S}}`

`\text{P(A)}=(4)/(10)`

`\text{P(B)}=\frac{\text{number of elements in B}}{\text{number of elements in S}}`

`\text{P(B)}=(5)/(10)`

`\text{P(A}\cap\text{B})=\frac{\text{number of elements in A and B}}{\text{number of elements in S}}`

`\text{P(A}\cap\text{B})=(2)/(10)`

Finally,

`\text{P(A}\cup\text{B})=(4)/(10)+(5)/(10)- (2)/(10)`

`\text{P(A}\cup\text{B})= (7)/(10)`

Answer:

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