Two dice are rolled simultaneously. Find the probability that the sum of the numbers on the faces is neither divisible by 3 nor by 4 .

Given: Two dice are thrown together. We know that, probability of occurrence of an event = Number of Desired outcomes Total number of outcomes . Let A denote the event of getting sum of the numbers on the two faces divisible by 3 , B denote the event of getting a sum of the numbers on the two faces divisible by 4 . Total outcomes are ( 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 ) Hence, total number of outcomes = 36 . Favourable outcomes is that sum of the numbers on the two faces in neither divisible by 3 nor by 4 . For getting sum of numbers on the two faces divisible by 3 . Favourable outcomes = ( 1 , 2 ) , ( 2 , 1 ) , ( 1 , 5 ) , ( 5 , 1 ) , ( 3 , 3 ) , ( 2 , 4 ) , ( 4 , 2 ) , ( 3 , 6 ) , ( 6 , 3 ) , ( 4 , 5 ) , ( 5 , 4 ) , ( 6 , 6 ) Favourable number of outcomes = 12 ∴ P ( A ) = 12 36 For getting sum of numbers on the two faces divisible by 4 . Favourable outcomes = ( 2 , 2 ) , ( 1 , 3 ) , ( 3 , 1 ) , ( 2 , 6 ) , ( 6 , 2 ) , ( 4 , 4 ) , ( 3 , 5 ) , ( 5 , 3 ) , ( 6 , 6 ) Favourable number of outcomes = 9 ∴ P ( B ) = 9 36 Probability of getting sum of the numbers on the two faces divisible by 3 and 4 = P ( A and B ) = 1 36 From addition theorem of probability, P ( A or B ) = P ( A ) + P ( B ) - P ( A and B ) Probability that the sum of the numbers on the two faces is neither divisible by 3 nor by 4 is P ( A ¯ ∩ B ¯ ) = P A ∪ B ¯ = 1 - P A ∪ B = 1 - 12 36 + 9 36 - 1 36 = 1 - 12 + 9 - 1 36 = 1 - 20 36 = 1 - 5 9 = 4 9 Therefore, the probability of getting the sum of the numbers on the two faces is neither divisible by 3 nor by 4 = 4 9 .

The standard deviation of 20 observations is 5 . If each observation is multiplied by 2 , find the standard deviation and variance of the resulting observations.

Let the observations be x 1 , x 2 , x 3 , . . . . . . . . . , x 20 and x ¯ be their mean. Variance = 5 and n = 20 Variance = 1 n ∑ x i - x 2 Therefore, 1 20 ∑ x i - x 2 = 5 ⇒ ∑ x i - x 2 = 20 5 _____________ 1 If each observation is multiplied by 2 , we get new observations. Let new observations be y 1 , y 2 , y 3 , . . . . . . . . . . . , y 20 . Where y i = 2 x i _________________ 2 We need to find variance of new observations. New variance = 1 n ∑ y i - y 2 Calculating y = 1 n ∑ y i ⇒ y = 1 20 ∑ 2 x i ⇒ y = 2 1 20 ∑ x i ⇒ y = 2 x _____________ 3 Equation 1 ⇒ ∑ x i - x 2 = 20 5 ⇒ ∑ 1 2 y i - 1 2 y 2 = 20 5 [From 2 and 3 ] ⇒ 1 2 2 ∑ y i - y 2 = 20 5 ⇒ ∑ y i - y 2 = 80 5 New variance = 1 n ∑ y i - y 2 = 80 5 × 1 20 = 4 5 Therefore, the required variance is 4 5 .