When two fair dice thrown together find the probability both dice show even numbers

Answer

Verified

Hint: Calculate the number of possible outcomes for throwing two dice. Calculate the number of favourable outcomes for each of the cases. Use the fact that the probability of any event is the ratio of the number of favourable outcomes and the number of possible outcomes to calculate the probability of each of the events.Complete step-by-step solution -
We have to calculate the probability of each of the events when two dice are thrown.
We know that the total number of possible outcomes when two dice are thrown is $=6\times 6=36$.
We know that the probability of any event is the ratio of the number of favourable outcomes and the number of possible outcomes.
We will now calculate the probability of events in each case.
(a) We have to calculate the probability that the sum of digits is a prime number.
We will draw a table showing the sum of digits on rolling both the dice.

+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

We observe that the possible values of prime numbers when two digits on the dice are added are 2, 3, 5, 7, and 11.
We observe that 2 occurs only once, 3 occurs 2 times, 5 occurs 4 times, 7 occurs 6 times and 11 occurs 2 times.
The number of favourable outcomes is the sum of occurrences of all the favourable outcomes. So, the number of favourable outcomes $=1+2+4+6+2=15$.
We know that the number of possible outcomes is 36.
Thus, the probability of getting the sum of two numbers as prime numbers is $=\dfrac{15}{36}=\dfrac{5}{12}$.
(b) We will now calculate the probability of occurrence of a doublet of an even number.
We know that the favourable outcomes are (2, 2), (4, 4), and (6, 6).
So, the number of favourable outcomes is 3.
We know that the number of possible outcomes is 36.
Thus, the probability of getting a doublet of an even number is $=\dfrac{3}{36}=\dfrac{1}{12}$.
(c) We will calculate the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice.
Possible multiples of 2 on dice are 2, 4, and 6.
Possible multiples of 3 on dice are 3 and 6.
The possible outcomes for multiples of 2 on one dice and multiple of 3 on other dice are (2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (6, 2), (3, 4), (6, 4), and (3, 6).
So, the number of favourable outcomes is 11.
We know that the number of possible outcomes is 36.
Thus, the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice is $=\dfrac{11}{36}$.
(d) We will calculate the probability of getting a multiple of 3 as a sum of digits on both the dice.
We will draw the table showing possible values of the sum of digits on both the dice.

+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

The possible values of multiples of 3 as a sum of digits on dice are 3, 6, 9, and 12.
We observe that 3 occurs 2 times, 6 occurs 5 times, 9 occurs 4 times and 12 occurs once.
The number of favourable outcomes is the sum of occurrences of all the favourable outcomes. So, the number of favourable outcomes $=2+5+4+1=12$.
We know that the number of possible outcomes is 36.
Thus, the probability of getting multiples of 3 as a sum of digits on dice is $=\dfrac{12}{36}=\dfrac{1}{3}$.
Note: We must calculate the number of favourable and possible outcomes in each case to calculate the probability of each of the given events. We should also be careful that we don’t count the same event repeatedly or we miss some event.

Try out challenging quizzes on this topic
made by experts!

Take me there!

When rolling two dice, distinguish between them in some way: a first one and second one, a left and a right, a red and a green, etc. Let (a,b) denote a possible outcome of rolling the two die, with a the number on the top of the first die and b the number on the top of the second die. Note that each of a and b can be any of the integers from 1 through 6. Here is a listing of all the joint possibilities for (a,b):

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

Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36. The set of all possible outcomes for (a,b) is called the sample space of this probability experiment.

With the sample space now identified, formal probability theory requires that we identify the possible events. These are always subsets of the sample space, and must form a sigma-algebra. In an example such as this, where the sample space is finite because it has only 36 different outcomes, it is perhaps easiest to simply declare ALL subsets of the sample space to be possible events. That will be a sigma-algebra and avoids what might otherwise be an annoying technical difficulty. We make that declaration with this example of two dice.

With the above declaration, the outcomes where the sum of the two dice is equal to 5 form an event. If we call this event E, we have

E={(1,4),(2,3),(3,2),(4,1)}.

Note that we have listed all the ways a first die and second die add up to 5 when we look at their top faces.

Consider next the probability of E, P(E). Here we need more information. If the two dice are fair and independent , each possibility (a,b) is equally likely. Because there are 36 possibilities in all, and the sum of their probabilities must equal 1, each singleton event {(a,b)} is assigned probability equal to 1/36. Because E is composed of 4 such distinct singleton events, P(E)=4/36= 1/9.

In general, when the two dice are fair and independent, the probability of any event is the number of elements in the event divided by 36.

What if the dice aren't fair, or aren't independent of each other? Then each outcome {(a,b)} is assigned a probability (a number in [0,1]) whose sum over all 36 outcomes is equal to 1. These probabilities aren't all equal, and must be estimated by experiment or inferred from other hypotheses about how the dice are related and and how likely each number is on each of the dice. Then the probability of an event such as E is the sum of the probabilities of the singleton events {(a,b)} that make up E.