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- The sum of the two numbers on two dice is called the score on two dice.
- The minimum score on two dice is 2 and the maximum score on two dice is 12.
- The cases favourable to a particular score can be read along the diagonal of that score.

#### Example – 01:

- Two fair dice are tossed. Find the probability that
- Solution:

- Two fair dice are thrown
- The sample space is

S = {1,2,3,4,5,6} × {1,2,3,4,5,6}

S | = | { | (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) | } |

∴ n (S) = 36

**the sum of the scores is even**

Let A be the event of that the sum of the scores is even i.e. 2, 4, 6, 8, 10, 12

∴ A = { (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) }

∴ n(A) = 18

By the definition P(A) = n(A)/n(S) = 18/36 = 1/2

Therefore the probability that the sum of the scores is even is 1/2

**the sum of the scores is odd**

Let B be the event of that the sum of the scores is odd i.e. 3, 5, 7, 9, 11

∴ B = { (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6),

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

(3, 6), (4, 5), (5, 4), (6, 3), (6, 5), (5, 6) }

∴ n(A) = 18

By the definition P(B) = n(B)/n(S) = 18/36 = 1/2

Therefore the probability that the sum of the scores is odd is 1/2

**the sum of the scores is a perfect square**

Let C be the event of that the sum of the scores is a perfect square i.e. 4, 9.

∴ C = { (1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3) }

∴ n(C) = 7

By the definition P(C) = n(C)/n(S) = 7/36

Therefore the probability that the sum of the scores is a perfect square is 7/36

**the sum of the score is a multiple of four or the sum of square is divisible by 4**

Let D be the event of that the sum of the score is a multiple of four i.e. 4, 8, 12

∴ D = { (1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6) }

∴ n(D) = 9

By the definition P(D) = n(D)/n(S) = 9/36 = 1/4

Therefore the probability that the sum of the score is a multiple of four is 1/4

**the sum of the scores is a multiple of 3**

Let E be the event of that the sum of the scores is a multiple of 3 i.e. 3, 6, 9, 12

∴ E = { (1, 2), (2, 1), (1, 5), (2, 4), (3, 3), (4, 2),

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

∴ n(E) = 12

By the definition P(E) = n(E)/n(S) = 12/36 = 1/3

Therefore the probability that the sum of the score is a multiple of 3 is 1/3..

**the sum of the points obtained is greater than 4**

Let F be the event of that the sum of the points obtained is

greater than 4 i.e. 5, 6, 7, 8, 9, 10, 11, 12.

∴ F = { (1, 4), (1, 5), (1, 6) (2. 3), (2, 4), (2, 5). (2, 6), (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, 1I), (6, 2), (6, 3). (6, 4), (6, 5), (6, 6) }

∴ n(F) = 30

By the definition P(F) = n(F)/n(S) = 30/36 = 5/6

Therefore the probability that the sum of the points obtained is greater than 4 is 5/6

**the sum of the points is at least 11**

Let G be the event of that the sum of the points is at least 11 i.e. 11, 12

∴ G ={ (5, 6), (6, 5), (6, 6) }

∴ n(G) = 3

By the definition P(G) = n(G)/n(S) = 3/36 = 1/12

Therefore the probability that the sum of the points is at least 11 is 1/12.

**the same score on the first die and second die**

Let H be the event of that the same score on the first die and second die.

∴ H = { (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) }

∴ n(H) = 6

By the definition P(H) = n(H)/n(S) = 6/36 = 1/6

Therefore the probability that the same score on the first die and second die is 1/6.

**the score on the second die is greater than the score on the first die**

Let J be the event of that the score on the second die is greater than the score on the first die

∴ J = { (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2. 3), (2, 4), (2, 5).

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

∴ n(J) = 15

By the definition P(J) = n(J)/n(S) = 15/36 = 5/12

Therefore the probability that the score on the second die is greater than the score on the first die is 5/12.

**the sum of the numbers on their faces obtained is either a perfect square or their sum is less than 5.**

Let K be the event of that the sum of the numbers on their faces

obtained is either a perfect square or their sum is less than 5.

∴ K = { (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3. 1), (3, 6), (4, 5), (5, 4), (6, 3) }

∴ n(K) = 10

By the definition P(K) = n(K)/n(S) = 10/36 = 5/18

Therefore the probability that the sum of the numbers on their faces

obtained is either a perfect square or their sum is less than 5 is 5/18.

**the sum of numbers shown is 7 or product is 12**

Let M be the event of that the sum of numbers shown is 7 or product is 12

∴ M = { (1, 6), (2, 5), (2, 6), (3, 4), (4, 3), (5, 2), (6, 1), (6, 2) }

∴ n(M) = 8

By the definition P(M) = n(M)/n(S) = 8/36 = 2/9

Therefore the probability that the sum of numbers shown is 7 or product is 12 is 2/9.

**the sum of these scores is either a perfect square or a prime number**

Let N be the event of that the sum of these scores is either

a perfect square or a prime number i.e. 4, 9, 2, 3, 5, 7, 11

∴ N = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3), (1. 1), (1, 2),

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

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

∴ n(N) = 22

By the definition P(N) = n(N)/n(S) = 22/36 = 11/18

Therefore the probability that the sum of numbers a perfect square or a prime number is 11/18.

**the product of the scores is 12**

Let Q be the event of that the product of the scores is 12

∴ Q = { (2, 6), (3, 4), (4, 3), (6, 2) }

∴ n(Q) = 4

By the definition P(Q) = n(Q)/n(S) = 4/36 = 1/9

Therefore the probability that the product of the scores is 12 is 1/9

**the sum of these scores is either a perfect square or an even number**

Let R be the event of that the sum of these scores is either

a perfect square or an even number i.e. 4, 9, 2, 6, 8,10, 12

∴ R = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3), (1. 1), (1, 5),

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

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

∴ n(R) = 22

By the definition P(R) = n(R)/n(S) = 22/36 = 11/18

Therefore the probability that the sum of these scores is either a perfect square or an even number is 11/18.

**the sum of these scores is either an even number or a number divisible by 5**

Let T be the event of that the sum of these scores is either

an even number or a number divisible by 5 i.e. 2, 4, 6, 8,10, 12, 5

∴ T = {(1. 1), (1, 3), (2, 2), (3, 1), (1, 5), (2, 4), (3, 3),

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

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

∴ n(T) = 22

By the definition P(T) = n(T)/n(S) = 22/36 = 11/18

Therefore the probability that the sum of these scores is either

an even number or a number divisible by 5 is 11/18.

**the sum of these scores is either a perfect square or a multiple of 3**

Let T be the event of that the sum of these scores is either

a perfect square or a multiple of 3 i.e. 4, 9, 3, 6, 12

∴ R = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3), (1. 2), (2, 1),

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

∴ n(T) = 14

By the definition P(T) = n(T)/n(S) = 14/36 = 7/18

Therefore the probability that the sum of these scores is either

a perfect square or a multiple of 3 is 7/18.

**the sum of the scores is either greater than 9 and an even number.**

Let U be the event of that the sum of these scores is either

greater than 9 i.e. 10, 12

∴ U = {(4, 6), (5, 5), (6, 4), (6, 6) }

∴ n(U) = 4

By the definition P(U) = n(U)/n(S) = 4/36 = 1/9

Therefore the probability that the sum of these scores is either greater than 9 and an even number is 1/4.

**the product of the scores is a perfect square**

Let V be the event of that the product of the scores is

a perfect square. i.e. 1, 4, 9, 16, 25, 36

∴ V = {(1, 1), (1, 4), (2, 2), (4, 1), (3, 3), (4, 4), (5, 5), (6, 6) }

∴ n(V) = 8

By the definition P(V) = n(V)/n(S) = 8/36 = 2/9

Therefore the probability that the product of the scores is a perfect square. is 2/9.

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