Question

Jason and Kyle both pick a number from 1 to 12 at random. What is the probability of both numbers being odd?

Answer 1

`\begin{gathered} 1\text{ - odd} \\ 2\text{ - } \\ 3\text{ - odd} \\ 4\text{ - } \\ 5\text{ - odd} \\ 6- \\ 7\text{ - odd} \\ 8\text{ - } \\ 9\text{ - odd} \\ 10\text{ - } \\ 11\text{ - odd} \\ 12 \end{gathered}`

We have a total of 12 numbers, where 6 of them are odd, then, the probability of picking an odd number is

`P(\text{odd\rparen = 6/12 = 1/2}`

The probability of an event A and B is

`P(A\text{ and }B)=P(A)\cdot P(B)`

Therefore

`\begin{gathered} P(\text{odd and odd})=P(\text{odd})\cdot P(\text{odd}) \\ \\ P(\text{odd and odd})=(1)/(2)\cdot(1)/(2) \\ \\ P(\text{odd and odd})=(1)/(4) \end{gathered}`

The probability of both numbers being odd is 1/4