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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 as A = (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.

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 as
A = (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.

Grade:10

1 Answers

Jitender Pal
askIITians Faculty 365 Points
9 years ago
Hello Student,
Please find the answer to your question
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.

Thanks
Jitender Pal
askIITians Faculty

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