If A and B are two mutually non-exclusive events such that P ( A ) = 1 4 , P ( B ) = 2 5 and P ( A ∪ B ) = 1 2 . Find the values of P ( A ∩ B ) and P A ∩ B c .

Given P ( A ) = 1 4 , P ( B ) = 2 5 , P ( A ∪ B ) = 1 2 . To find: , P ( A ∩ B ) = ? and P A ∩ B c = ? ∵ P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) ⇒ 1 2 = 1 4 + 2 5 - P ( A ∩ B ) ∴ P ( A ∩ B ) = 1 4 + 2 5 - 1 2 = 5 + 8 - 10 20 = 3 20 P A ∩ B C = P ( A ) - P ( A ∩ B ) = 1 4 - 3 20 ∴ P A ∩ B c = 5 - 3 20 = 1 10

The diameters of circles (in mm) drawn in a design are given below, calculate variance [Use short-cut method] Diameter Number of Circles 33 - 36 15 37 - 40 17 41 - 44 21 45 - 48 22 49 - 52 25

Class Modify class Mid-point f i d i = x i - a h d i 2 f i d i f i d i 2 33 - 36 32 · 5 - 36 . 5 34 . 5 15 - 2 4 - 30 60 37 - 40 36 · 5 - 40 · 5 38 · 5 17 - 1 1 - 17 17 41 - 44 40 · 5 - 44 - 5 42 . 5 21 0 0 0 0 45 - 48 44 · 5 - 48 . 5 46 · 5 22 1 1 22 22 49 - 52 48 - 5 - 52 · 5 50 . 5 25 2 4 50 100 Σ f i = N = 100 Σ f i d i = 25 Σ f i d i 2 = 199 Let the assumed mean a = 42 . 5 Therefore, mean = a + Σ f i d i N × h , here h = class width i.e., h = 4 = 42 · 5 + 25 100 × 4 = 42 · 5 + 25 25 = 42 · 5 + 1 = 43 · 5 Thus, variance = h 2 1 N · Σ f i d i 2 - Σ f i d i N 2 = ( 4 ) 2 199 100 - 25 100 2 = 31 · 84 - 1 = 30 · 84 Hence, the required variance = 30 · 84 .

The Probabilities of three events A , B , C are such that P ( A ) = 0 . 3 , P ( B ) = 0 . 4 , P ( C ) = 0 . 8 , P ( A ∩ B ) = 0 . 09 , P ( A ∩ C ) = 0 . 28 , P ( A ∩ B ∩ C ) = 0 . 08 and P ( A ∪ B ∪ C ) ≥ 0 . 75 . Show that P ( B ∩ C ) ) lies in the interval 0 . 21 , 0 . 46 .

Given: P ( A ∪ B ∪ C ) ⩾ 0 . 75 ⇒ 0 . 75 ⩽ P ( A ∪ B ∪ C ) ⩽ 1 Since, probability of sample space is one i.e P ( s ) = 1 and here, P ( A ∪ B ∪ C ) ⩽ 1 . We know that, from Addition Theorem of Probability for Three Events P A ∪ B ∪ C = P ( A ) + P ( B ) + P ( C ) - P ( A ∩ B ) - P ( B ∩ C ) - P ( C ∩ A ) + P ( A ∩ B ∩ C ) Now, 0 . 75 ⩽ P ( A ) + P ( B ) + P ( C ) - P ( A ∩ B ) - P ( B ∩ C ) - P ( C ∩ A ) + P ( A ∩ B ∩ C ) ⩽ 1 ⇒ 0 . 75 ⩽ 0 . 3 + 0 . 4 + 0 . 8 - 0 . 09 - 0 . 28 - P ( B ∩ C ) + 0 . 08 ⩽ 1 ⇒ 0 . 75 ⩽ 1 . 21 - P ( B ∩ C ) ⩽ 1 ⇒ - 0 . 46 ⩽ - P ( B ∩ C ) ⩽ - 0 . 21 ⇒ 0 . 46 ⩾ P ( B ∩ C ) ⩾ 0 . 21 ∴ P ( B ∩ C ) lies in the interval [ 0 . 21 , 0 . 46 ] . Hence, showed.

The probabilities of three events A , B , C are such that P ( A ) = 0 . 3 , P ( B ) = 0 . 4 , P ( C ) = 0 . 8 , P ( A ∩ B ) = 0 . 08 , P ( A ∩ C ) = 0 . 28 , P ( A ∩ B ∩ C ) = 0 . 09 , and P ( A ∪ B ∪ C ) ≥ 0 . 75 , show that P ( B ∩ C ) lies in the interval [ 0 . 23 , 0 . 48 ] .

Given, P ( A ) = 0 . 3 , P ( B ) = 0 . 4 , P ( C ) = 0 . 8 , P ( A ∩ B ) = 0 . 08 , P ( A ∩ C ) = 0 . 28 , P ( A ∩ B ∩ C ) = 0 . 09 Since, P ( A ∪ B ∪ C ) ≥ 0 . 75 ⇒ 0 . 75 ≤ P ( A ∪ B ∪ C ) ≤ 1 ⇒ 0 . 75 ≤ P ( A ) + P ( B ) + P ( C ) - P ( A ∩ B ) - P ( B ∩ C ) - P ( A ∩ C ) + P A ∩ B ∩ C ≤ 1 ⇒ 0 . 75 ≤ 0 . 3 + 0 . 4 + 0 . 8 - 0 . 08 - P ( B ∩ C ) - 0 . 28 + 0 . 09 ≤ 1 ⇒ 0 . 75 ≤ 1 . 23 - p ( B ∩ C ) ≤ 1 ⇒ - 0 . 48 ≤ - P ( B ∩ C ) ≤ - 0 . 23 ⇒ 0 . 23 ≤ P ( B ∩ C ) ≤ 0 . 48 Hence, proved.

Given that P ( A ) = 0 . 35 , P ( B ) = 0 . 73 and P ( A ∩ B ) = 0 . 14 , find P ( A ∪ B )

Given that, P ( A ) = 0 . 35 , P ( B ) = 0 . 73 and P ( A ∩ B ) = 0 . 14 By De Morgan's law P ( A ∩ B ) = 1 - P ( A ∪ B ) Hence, P ( A ∪ B ) = 1 - P ( A ∩ B ) = 1 - 0 . 14 = 0 . 86

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 .