2. 15In the experiment of throwing two fair dice, let A be the event that the fi

Question

2. 15In the experiment of throwing two fair dice, let A be the event that the first die is odd, B be the event that the second die is odd, and C be the event that the sum is odd (a) Describe the events A, B, and C using set notation. (b) Are events A and B independent? Are events B and C independent? Justify your answers (c) Are events A, B, and C independent? Justify your answer 3. 10 Ten percent of items from a certain production line are defective. What is the probability that there is more than one defective item in a batch of 100 items?

Explanation / Answer

Q.2 We have to throw two dice let say D1 is first die and D2 is second die

A = First Die is odd (D1 = 1,3,5)

B = Second die is odd ( D2 = 1,3,5)

C = Sum is odd (D1 + D2 = 3, 5, 7, 9, 11)

(a) A = {(1,1), (1,2) , (1,3) , (1,4) , (1,5) , (1,6), (3,1), (3,2) , (3,3) , (3,4) , (3,5) , (3,6). (5,1), (5,2) , (5,3) , (5,4) , (5,5) , (5,6)}

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

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

(b) A and B are independent

P(A B) = P(A) P(B)

P(A B) = 9/36 = 1/4

P(A) = 18/36 and P(B) = 18/36

P(A) P(B) = 1/4

so yes both events are independent.

Are B or C is independent

so P(B C) = P(B)P(C)

P(B) = 1/2 ; P(C) = 1/2

P(B C) = 9/36 = 1/4

so yes, B and C are also independent.

(c) P(A B C) = P(A) × P(B) × P(C)

so P(A B C) = 0 because sum can't be good if both A and B are odd.

so A, B and C are not independent.

Q.3 Pr (Defective) = 10% = 0.1

Expeccted number of items to be defective = 100 * 0.1 = 10

As n = 100, which is large so we can do normal approximation

Standard deviation of numbe of defective items out of 100 = sqrt [ 0.1 * 0.9 * 100] = 3

Pr(More then one defective item in a batch of 100 items) = 1 - Pr ( only zero or one defective item)

= 1 - Pr (X <=1 ; 10 ; 3)

Z= (1 -10)/3 = -3

so Pr(More then one defective item in a batch of 100 items) = 1 - (-3) = 1 - 0.0013 = 0.9987

so there are 99.87% probability that there is more than one defective item in a batch of 100 items.

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