A lot contains 50 defective and 50 non-defectivebulbs. Two bulbs are drawn at random one at a time withreplacementevents A, and as first bul. The B C

Question

A lot contains 50 defective and 50 non-defectivebulbs. Two bulbs are drawn at random one at a time withreplacementevents A, and as first bul. The B C

A lot contains 50 defective and 50 non-defectivebulbs. Two bulbs are drawn at random one at a time withreplacementevents A, and as first bul. The B C are defined theis defective, the second bulb is non-defective, the two banboth defective or non-defective, respectively. Then,(a) A, B and C are pairwise independent(b) A, B and C are pairwise not independent(c) A, B and C are independent(d) None of the above
A. A and B are independent
B. B and C are independent
C. A and C are independent
D. A,B and C are pairwise independent

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - A::B::C
Let X = defective and Y = non-defective. Then all possible outcomes are `{XX,XY,YX,YY}`
Aslo, `P(XX)=(50)/(100)xx(50)/(100)=1/4`
`P(XY)=(50)/(100)xx(50)/(100)=1/4`
`P(YX)=(50)/(100)xx(50)/(100)=1/4`
`P(YY)=(50)/(100)xx(50)/(100)=1/4`
Here, `A=(XX)uu(XY),B=(XY)uu(YY),C=(XX)uu(YY)`
`thereforeP(A)=P(XX)+P(XY)=1/4+1/4=1/2`
`thereforeP(B)=P(XY)+P(YX)=1/4+1/4=1/2`
Now, `P(AxxB)=P(XY)=1/4=P(A)xxP(B)`
Thus, A and B are independent events.
`P(CxxA)=P(XY)=1/4=P(C)xxP(A)`
Thus,C and A are independent events.
`P(AxxBxxC)=0` (impossible events)
`ne P(A)xxP(B)xxP(C)`
Thus, A, B and C are depended events.
Therefore, we can conclude that A, B, C are pairwise independent but A,B ,C are dependent.