What is the probability of getting a sum that is multiples of three when two dice are thrown

n(s) = 36 i.e.

(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)}

Event = {multiple of 3 as a sum}= {(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (3, 6), (6, 3), (5, 4), (4, 5), (6, 6)}n(E) = 12

p(E) = ?

∴ P(E) = `"n(E)"/"n(S)" = 12/36 = 1/3`.

Text Solution

Solution : Desired sums of the numbers are 3, 6, 9 and 12,<br>n (S) = 2 + 5 + 4 + 1<brge 12<br> `therefore`Reqd. probability=`12/36=1/3`

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.

When a die is rolled (or thrown), we get the outcomes as:

{1, 2, 3, 4, 5, 6}

That means, the total number of outcomes = n(S) = 6

Let E be the event of getting a multiple of 3.

E = {3, 6}

Number of favorable outcomes of the event E = n(E) = 2

P(E) = n(E)/n(S) = 2/6 = ⅓

Hence, the probability of getting a multiple of 3 when a die is thrown is 1/3.

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What is the probability that the sum of $2$ dice rolls is a multiple of $3$? What about for $3$ dice rolls? For $n$ dice rolls?

So I have the first part of this solution worked out by writing out all the combinations of $2$ dice rolls (I won't write them here). There are $6^2=36$ total and the following multiples of $3$ are rolled: $$3\rightarrow2 ways$$$$6\rightarrow5ways$$$$9\rightarrow4ways$$$$12\rightarrow1way$$So adding up all the ways over the total rolls is the probability:$$\frac{2+5+4+1}{6^2}=\frac{12}{36}=\frac{1}{3}=0.33$$The second part and the general solution is what trips me up. I know there are $6^3$ possible rolls for $3$ dice, and we're now including the multiples $15$ and $18$, but how do I figure out how many ways there are to roll each multiple without counting all $216$ possibilities? How would I apply this to a general solution with $n$ dice?