Question

If you flip a coin and roll a die, what is the probability that you will flip a head and roll a 3?

Answer 1

Hint: We use the concepts of probability to solve this problem. We will learn about all the conditions that will occur when a die is rolled and when a coin is flipped. We will know in detail about probability and its properties. We will also discuss another way to solve this problem.

Complete step-by-step solution:
In mathematics, probability is defined as the occurrence of a random event. It is also defined as the ratio of number of favorable outcomes to total number of outcomes. The probability of any event always lies in the range \[[0,1]\].
So, \[P(E) = \dfrac{{{\text{number of favorable outcomes}}}}{{{\text{total number of outcomes}}{\text{.}}}}\] is the probability of an event E.
If the probability of an event is 0, then the event doesn’t happen.
If the probability of an event is 1, then it will happen for sure.
So, now, when a coin is flipped, then there are two outcomes which are getting a head or a tail.
And when a die is rolled, there are six outcomes which are, getting 1, 2, 3, 4, 5 and 6 on the top face of the die.
Let the event of getting a head on coin and 3 on die be E.
So, when both the things happen at a time i.e., when a coin is flipped and a die is rolled simultaneously, then the total possible outcomes are
\[(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)\]
So, there are 12 possible outcomes and the favourable outcome (the outcome that we need) is \[(H,3)\]
So, probability of getting a head on coin and 3 on die is \[P(E) = \dfrac{{{\text{number of favorable outcomes}}}}{{{\text{total number of outcomes}}{\text{.}}}}\]
Which is equal to \[P(E) = \dfrac{1}{{12}}\]

Note: We can also use another formula. In this, probability of an event G happening after the happening of another event E is given by \[P(E,G) = P(E) \times P(G)\]
So, here, let probability of getting head be \[P(H)\] and probability of getting a 3 on die be \[P(D)\]
So, probability of getting head and 3 on die is \[P(H,D) = P(H) \times P(D)\]
\[ \Rightarrow P(H,D) = \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{{12}}\]
So, we can also calculate probability in this way.