Question

Two different dice are thrown together, Find the probability that the numbers obtained have
a) Even sum, and
b) Even product

Answer 1

Hint: We need to use the definition of probability to solve this question which is $P(E)=\dfrac{n(E)}{n(S)}$ where n(E) is the number of elements in the Event Set and n(S) is the number of elements in the Sample Space. Only by using this definition we will be able to solve this question.

Complete step by step answer:
First of all we need to understand what is ‘Event’ and what is ‘Sample Space’. Event is the set of all favourable outcomes or we can say Event is the subset of sample space whereas Sample Space is the set of all possible outcomes of an event.

First of all let us write the sample space when two different dice are thrown together.
$S=\left\{ \begin{align}

  & \left( 1,1 \right),(1,2),..........(1,6) \\

 & (2,1),(2,2),.........(2,6) \\

 & (3,1),(3,2),..........(3,6) \\

 & (4,1),(4,2),.........(4,6) \\

 & (5,1),(5,2),..........(5,6) \\

 & (6,1),(6,2)...........(6,6) \\

\end{align} \right\}$

The elements in this set is $6\times 6=36$ , therefore n(S)=36.

Where the ordered pair (a,b) represents the number on the first and second die.
(a)

Now let us write the Event Set for the event: The numbers obtained have an even sum. As we know sum of two even numbers is an even number and sum of two odd numbers is also an even number but sum of an even and an odd number is not an even number therefore we have,

$E=\left\{ \begin{align}

  & (1,1),(1,3),(1,5) \\

 & (2,2),(2,4),(2,6) \\

 & (3,1),(3,3),(3,5) \\

 & (4,2),(4,4),(4,6) \\

 & (5,1),(5,3),(5,5) \\

 & (6,2),(6,4),(6,6) \\

\end{align} \right\}$

The number of elements in this set is $6\times 3=18$ , therefore we have n(E)=18

Hence, probability of this event is given by, $P(E)=\dfrac{n(E)}{n(S)}$ .

Substituting the value of n€ and n(S) we get,

$P(E)=\dfrac{18}{36}\Rightarrow P(E)=\dfrac{1}{2}$

Hence, the answer is $\dfrac{1}{2}$ .

(b)

Now we write the event set for the event: The numbers obtained have an even product.

As we know two numbers can have an even product only when one of them is even or both of them are even. Hence, we can write

$E=\left\{ \begin{align}

  & (1,2),(1,4),(1,6) \\

 & (2,1),(2,2),......(2,6) \\

 & (3,2),(3,4),(3,6) \\

 & (4,1),(4,2),......(4,6) \\

 & (5,2),(5,4),(5,6) \\

 & (6,1),(6,2),......(6,6) \\

\end{align} \right\}$

The number of elements in this set is $3\times 3+3\times 6=9+18=27$ . Therefore, n(E)=27
Hence, probability is given by, $P(E)=\dfrac{n(E)}{n(S)}$

Substituting n(E) and n(S) we get, $P(E)=\dfrac{27}{36}\Rightarrow P(E)=\dfrac{3}{4}$
Hence, the probability is $\dfrac{3}{4}$ .


Note: As we can see we wrote the Sample Space first and used it for both because when dice are thrown then the possible outcomes are the same for both the events. We had mentioned earlier that an event is a subset of sample space. We can also verify this by looking at both Event sets in our question.