Given that a coin is tossed and a die is rolled.
Let's find the probability of getting a head and a number greater than 1.
To find the probability of getting a head and a number greater than 1, apply the formula:
P(head and 1) = P(head) x P(greater than 1)
Where:
Number of heads in a coin = 1
A coin contains a head and a tail. ==> 2 possible outcomes
Numbers greater than one in a die = 2, 3, 4, 5, 6, ==> 5 numbers.
A dies contains 6 numbers ==> 6 possible outcomes
Therefore, we have:
![\begin{gathered} P(\text{head)}=(1)/(2) \\ \\ P(\text{greater than 1)=}(5)/(6) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/l40njcyhy6n807k5wy8nu8v01hvf0aunov.png)
Thus, finding the probability, we now have:
![\begin{gathered} P(\text{head and 1) = }(1)/(2)*(5)/(6)=(1*5)/(2*6)=(5)/(12) \\ \\ P(\text{head and 1) = }(5)/(12) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mjp27ufocbbf7b7tg0p5eeoeykpptv179g.png)
Therefore, the probability of getting a head and a number greater than 1 is:
![(5)/(12)](https://img.qammunity.org/2023/formulas/mathematics/college/5jq9pt23sip8kqh58x3ldaxpwe1cwnko1j.png)
ANSWER:
![(5)/(12)](https://img.qammunity.org/2023/formulas/mathematics/college/5jq9pt23sip8kqh58x3ldaxpwe1cwnko1j.png)