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 a…

Question

asked Apr 6, 2024 228k views

1 Answer

Eric Miller · Answer 1 · 2024-04-12T14:12:15+0000

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