A lot contains 50 defective and 50 non defective bulbs. Two bulbs are

1.	A lot contains 50 defective and 50 non defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events A, B, C are defined asA = (the first bulb is defective)B = (the second bulb is non – defective)C = (the two bulbs are both defective or both non defective) Determine whether (i) A, B, C are pair wise independent (ii) A, B, C are independent.
Answer» Let S = defective and Y = non defective. Then all possible outcomes are {XX, XY, YX, YY} Also P (XX) = 50/100 x 50/100 = 1/4, P(XY) = 50/100 x 50/100 = 1/4, P(YX) = 50/100 x 50/100 = 1/4, P(YY) = 50/100 x 50/100 = 1/4 Here, A =XX ∪ XY; B = XY ∪ YY; C = XX ∪ YY ∴ P (A) = P(XX) + P(XY) = 1/4 + 1/4 = 1/2 ∴ P (B) = P (XY) + P (YX) = 1/4 + 1/4 = 1/2 P (C) = P (XX) + P (YY) = 1/4 + 1/4 =1/2 Now, P (AB) = P (XY) = 1/4 = P (A). P (B) ∴ A and B are independent events. P (BC) = P (YX) = 1/4 = P (B). P (C) ∴ B and C are independent events. P (CA) = P (XX) = 1/4 = P (C). P (A) ∴ C and A are independent events. P (ABC) = 0 (impossible event) ≠ P (A) P (B) P (C) ∴ A, B, C are dependent events, Thus we can conclude that A, B, C are pair wise independent bet A, B, C are dependent events.

A lot contains 50 defective and 50 non defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events A, B, C are defined asA = (the first bulb is defective)B = (the second bulb is non – defective)C = (the two bulbs are both defective or both non defective) Determine whether (i) A, B, C are pair wise independent (ii) A, B, C are independent.

Answer»

Let S = defective and Y = non defective. Then all possible outcomes are {XX, XY, YX, YY}

Also P (XX) = 50/100 x 50/100 = 1/4,

P(XY) = 50/100 x 50/100 = 1/4,

P(YX) = 50/100 x 50/100 = 1/4,

P(YY) = 50/100 x 50/100 = 1/4

Here, A =XX ∪ XY; B = XY ∪ YY; C = XX ∪ YY

∴ P (A) = P(XX) + P(XY) = 1/4 + 1/4 = 1/2

∴ P (B) = P (XY) + P (YX) = 1/4 + 1/4 = 1/2

P (C) = P (XX) + P (YY) = 1/4 + 1/4 =1/2

Now, P (AB) = P (XY) = 1/4 = P (A). P (B)

∴ A and B are independent events.

P (BC) = P (YX) = 1/4 = P (B). P (C)

∴ B and C are independent events.

P (CA) = P (XX) = 1/4 = P (C). P (A)

∴ C and A are independent events.

P (ABC) = 0 (impossible event)

≠ P (A) P (B) P (C)

∴ A, B, C are dependent events,

Thus we can conclude that A, B, C are pair wise independent bet A, B, C are dependent events.