Probability because that rolling 2 dice with the six sided dotssuch together 1, 2, 3, 4, 5 and also 6 dots in every die.

You are watching: Probability of rolling two dice and getting a sum of 7 or 11 When 2 dice are thrown simultaneously, thus variety of event have the right to be 62 = 36 due to the fact that each die has 1 come 6 number ~ above its faces. Climate the feasible outcomes are shown in the listed below table. Note:

(i) The outcomes (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) are referred to as doublets.

(ii) The pair (1, 2) and also (2, 1) are various outcomes.

Worked-out troubles involving probability because that rolling two dice:

1. two dice space rolled. Permit A, B, C be the events of acquiring a sum of 2, a amount of 3 and also a amount of 4 respectively. Then, present that

(i) A is a basic event

(ii) B and also C are compound events

(iii) A and also B space mutually exclusive

Solution:

Clearly, we haveA = (1, 1), B = (1, 2), (2, 1) and also C = (1, 3), (3, 1), (2, 2).

(i) since A consists of a solitary sample point, it is a simple event.

(ii) due to the fact that both B and also C contain an ext than one sample point, each one of them is a link event.

(iii) due to the fact that A ∩ B = ∅, A and B room mutually exclusive.

2. 2 dice space rolled. A is the event that the amount of the numbers shown on the 2 dice is 5, and B is the occasion that at the very least one that the dice shows up a 3. Are the two occasions (i) mutually exclusive, (ii) exhaustive? Give debates in assistance of her answer.

Solution:

When 2 dice room rolled, we have n(S) = (6 × 6) = 36.

Now, A = (1, 4), (2, 3), (4, 1), (3, 2), and

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

(i) A ∩ B = (2, 3), (3, 2) ≠ ∅.

Hence, A and B are not support exclusive.

(ii) Also, A ∪ B ≠ S.

Therefore, A and B are not exhaustive events.

More instances related to the inquiries on the probabilities for throwing two dice.

3. 2 dice are thrown simultaneously. Find the probability of:

(i) acquiring six together a product

(ii) acquiring sum ≤ 3

(iii) acquiring sum ≤ 10

(iv) obtaining a doublet

(v) gaining a sum of 8

(vi) gaining sum divisible through 5

(vii) getting sum of atleast 11

(viii) getting a lot of of 3 together the sum

(ix) gaining a full of atleast 10

(x) gaining an even number as the sum

(xi) getting a element number as the sum

(xii) acquiring a doublet of also numbers

(xiii) obtaining a multiple of 2 top top one die and also a many of 3 ~ above the various other die

Solution:

Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces. We know that in a solitary thrown the two different dice, the total number of possible outcomes is (6 × 6) = 36.

(i) getting six as a product:

Let E1 = occasion of gaining six together a product. The number who product is 6 will be E1 = <(1, 6), (2, 3), (3, 2), (6, 1)> = 4

Therefore, probability ofgetting ‘six together a product’

number of favorable outcomesP(E1) = Total number of possible result = 4/36 = 1/9

(ii) acquiring sum ≤ 3:

Let E2 = occasion of gaining sum ≤ 3. The number whose sum ≤ 3 will certainly be E2 = <(1, 1), (1, 2), (2, 1)> = 3

Therefore, probability ofgetting ‘sum ≤ 3’

number of favorable outcomesP(E2) = Total number of possible result = 3/36 = 1/12

(iii) obtaining sum ≤ 10:

Let E3 = event of acquiring sum ≤ 10. The number whose amount ≤ 10 will be E3 =

<(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),

(6, 1), (6, 2), (6, 3), (6, 4)> = 33

Therefore, probability ofgetting ‘sum ≤ 10’

variety of favorable outcomesP(E3) = Total variety of possible outcome = 33/36 = 11/12(iv)getting a doublet:Let E4 = event of getting a doublet. The number i m sorry doublet will certainly be E4 = <(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)> = 6

Therefore, probability ofgetting ‘a doublet’

number of favorable outcomesP(E4) = Total variety of possible result = 6/36 = 1/6

(v)getting a amount of 8:

Let E5 = event of gaining a sum of 8. The number i m sorry is a sum of 8 will certainly be E5 = <(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)> = 5

Therefore, probability ofgetting ‘a sum of 8’

number of favorable outcomesP(E5) = Total variety of possible result = 5/36

(vi)getting sum divisible through 5:

Let E6 = event of getting sum divisible through 5. The number whose amount divisible by 5 will be E6 = <(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)> = 7

Therefore, probability ofgetting ‘sum divisible by 5’

number of favorable outcomesP(E6) = Total variety of possible result = 7/36

(vii)getting sum of atleast 11:

Let E7 = event of gaining sum the atleast 11. The occasions of the sum of atleast 11 will certainly be E7 = <(5, 6), (6, 5), (6, 6)> = 3

Therefore, probability ofgetting ‘sum the atleast 11’

number of favorable outcomesP(E7) = Total number of possible outcome = 3/36 = 1/12

(viii) acquiring amultiple that 3 together the sum:

Let E8 = occasion of getting a many of 3 as the sum. The events of a lot of of 3 together the sum will it is in E8 = <(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3) (6, 6)> = 12

Therefore, probability ofgetting ‘a multiple of 3 as the sum’

variety of favorable outcomesP(E8) = Total variety of possible outcome = 12/36 = 1/3

(ix) acquiring a totalof atleast 10:

Let E9 = event of getting a total of atleast 10. The events of a complete of atleast 10 will certainly be E9 = <(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)> = 6

Therefore, probability ofgetting ‘a total of atleast 10’

variety of favorable outcomesP(E9) = Total number of possible outcome = 6/36 = 1/6

(x) acquiring an evennumber together the sum:

Let E10 = event of gaining an even number together the sum. The events of an even number as the amount will be E10 = <(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 3), (3, 1), (3, 5), (4, 4), (4, 2), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)> = 18

Therefore, probability ofgetting ‘an even number as the sum

variety of favorable outcomesP(E10) = Total number of possible result = 18/36 = 1/2

(xi) obtaining a primenumber together the sum:

Let E11 = occasion of gaining a element number together the sum. The occasions of a prime number as the sum will be E11 = <(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)> = 15

Therefore, probability ofgetting ‘a prime number as the sum’

variety of favorable outcomesP(E11) = Total number of possible result = 15/36 = 5/12

(xii) getting adoublet of also numbers:

Let E12 = event of acquiring a doublet of also numbers. The occasions of a double of also numbers will be E12 = <(2, 2), (4, 4), (6, 6)> = 3

Therefore, probability ofgetting ‘a doublet of even numbers’

variety of favorable outcomesP(E12) = Total number of possible outcome = 3/36 = 1/12

(xiii) gaining amultiple that 2 ~ above one die and a many of 3 on the other die:

Let E13 = event of getting a multiple of 2 top top one die and also a many of 3 ~ above the various other die. The events of a many of 2 on one die and also a lot of of 3 top top the other die will certainly be E13 = <(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)> = 11

Therefore, probability ofgetting ‘a many of 2 on one die and also a multiple of 3 on the other die’

number of favorable outcomesP(E13) = Total variety of possible outcome = 11/36

4. Twodice room thrown. Discover (i) the odds in favour of obtaining the sum 5, and also (ii) theodds versus getting the sum 6.

Solution:

We know that in a single thrown of 2 die, the complete numberof feasible outcomes is (6 × 6) = 36.

Let S be the sample space. Then,n(S) = 36.

(i) the odds in favour of acquiring the amount 5:

Let E1 it is in the event of acquiring the amount 5. Then,E1 = (1, 4), (2, 3), (3, 2), (4, 1) ⇒ P(E1) = 4Therefore, P(E1) = n(E1)/n(S) = 4/36 = 1/9⇒ odds in favour of E1 = P(E1)/<1 – P(E1)> = (1/9)/(1 – 1/9) = 1/8.

(ii) the odds versus getting the sum 6:

Let E2 be the event of obtaining the sum 6. Then,E2 = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) ⇒ P(E2) = 5Therefore, P(E2) = n(E2)/n(S) = 5/36 ⇒ odds against E2 = <1 – P(E2)>/P(E2) = (1 – 5/36)/(5/36) = 31/5.

See more: What Did Rutherford Observe That Surprised Him, What Did Rutherford'S Gold

5. Two dice, one blue and one orange, space rolled simultaneously. Uncover the probability of getting