30 points 3. Random variables X and Y have the following joint probability mass

Question

30 points 3. Random variables X and Y have the following joint probability mass function 0 0 0 0.14 0.01 0.12 0 0.02 0.03 0.13 0.14 0.15 And, it is known that EX+Y)-0 a. Find the appropriate values of a b. Find the value of E(X-2Y) and b 20 points 4. A new security system needs to be evaluated in the airport. The probability of a person being a security hazard is 4%. At the checkpoint, the security system denied a person without security problems 2% of the time. But the security system passed a person with security problems 1% of the time a. What is the probability that a person passes through the system? b. What is the probability that a person who passes through the system is without any security problems?

Explanation / Answer

30)

here for E(2X+Y)=2E(X)+E(Y)=2*(-2*(a+0.14+0.01+0.12)-1*(0+b+0.02+0.13)+0*(0+0+0.03+0.14)+1*(0+0+0+0.15))

+(0*(a+0+0+0)+1*(0.14+2+0+0)+2*(0.01+0.02+0.03+0)+3*(0.12+0.13+0.14+0.15)) =0

2(-2*(a+0.27)-(b+0.15)+0.15)+(0.14+b)+2*0.06+3*0.54 =0

-4a-1.08-2b-0.30+0.30+0.14+b+0.12+1.62=0

4a+b=0.8 ..................(1)

as sum of all probabilities =1

therefore a+0.14+0.01+0.12+b+0.02+0.13+0.03+0.14+0.15 =1

a+b=0.26 ................(2)

from above 2 equations:

a=0.18 ; b= 0.08

b) putting values of a and b in above table

E(X-Y) = E(X)-E(Y)=(-2*0.45-1*0.23+0*17+1*0.15)-(0*0.18+1*0.22+2*0.06+3*0.54)=-2.94

4)

probabilty that person passes through =P(person is security hazard and does passes through+person is not security hazard and does passes through)=0.04*0.01+(1-0.04)*(1-0.02)=0.9412

b)probability that person is without any security problems; given passes=P(person is not security hazard and does passes through)/P( person passes) =(1-0.04)*(1-0.02)/0.9412 =0.9996

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