Two different dice are thrown together what is the probability that the number on each dice is even

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)=18Hence, 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)=27Hence, 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.

Solution:The total number of outcomes when two dice are tossed together is 36.

The sample space is as follows

	1	2	3	4	5	6
1	(1,1)	(1,2)	(1,3)	(1,4)	(1,5)	(1,6)
2	(2,1)	(2,2)	(2,3)	(2,4)	(2,5)	(2,6)
3	(3,1)	(3,2)	(3,3)	(3,4)	(3,5)	(3,6)
4	(4,1)	(4,2)	(4,3)	(4,4)	(4,5)	(4,6)
5	(5,1)	(5,2)	(5,3)	(5,4)	(5,5)	(5,6)
6	(6,1)	(6,2)	(6,3)	(6,4)	(6,5)	(6,6)

i. Let E be the event ‘that number of each die is even’Favourable outcomes = { (2,2) , (2,4), (2,6), (4,2), (4,4), (4,6), (6,2),(6,4), (6,6) }

Probability that the number on each dice is even P(E)

`="Number of favourable outcomes"/"Total number of outcomes"=9/36=1/4`

ii. Let F be the eventiii. Favourable outcomes = { (1,4) (2,3) (3,2) (4,1) }

Probability that the sum of the numbers appearing on the two dice is 5

`="Number of favourable outcomes"/"Total number of outcomes"4/36=1/ 9`

Two different dice are tossed together. Find the probability thati the number on each die is even,ii the sum of the numbers appearing on the two dice is 5 .

Solution

When two different dice are thrown, then total number of outcomes = 36.
(i) Let E1 be the event of getting an even number on each die.
These numbers are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4) and (6, 6). Number of Favourable outcomes = 9
∴ P(getting an even number on both dice) = P(E1)=Number of outcomes favourable to E1Number of all possible outcomes

= 936=14

Thus, the probability of getting an even number on both dice is 14.

(ii) Let E2 be the event of getting the sum of the numbers appearing on the two dice is 5.
These numbers are (1, 4), (2, 3), (3, 2) and (4, 1) Number of Favourable outcomes = 4
∴ P(getting the sum of the numbers appearing on the two dice is 5) = P(E2)=Number of outcomes favourable to E2Number of all possible outcomes

= 436=19

Thus, the probability of getting the sum of the numbers appearing on the two dice is 5 is 19.

Mathematics

RS Aggarwal (2018)
Standard X

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Two different dice are tossed together. Find he probability that i the number on each die is even, ii the sum of the numbers appearing on the two dice is 5.

Solution