Does someone knows how to solve this problem in a fast way?

Hi Marcos,

Since X, Y, Z have only 1 or 0, 0<= XY <=1 and 0<= Z <= 1.

So we have 0<= XY + Z <= 2.

Therefore we have P(XY+Z=0) +P(XY+Z=1) +P(XY+Z=2) =1.

That implies P(XY+Z=0) = 1-(P(XY+Z=1) +P(XY+Z=2))

= 1 - P(XY + Z >= 1)

= 1 - 0,55 = 0,45


Now the case XY+Z = 0 can be splitted into:

X=0 and Y=0 and Z=0

X=0 and Y=1 and Z=0

X=1 and Y=0 and Z=0

So we have P(XY + Z =0) = P(X=0)*P(Y=0)*P(Z=0) + P(X=0)*P(Y=1)*P(Z=0) + P(X=1)*P(Y=0)*P(Z=0)

( because X, Y, Z are independent).


Let P(X=1)=a and P(Y=1)=b. And P(Z=0)= 1- P(Z=1) =3/4.

Therefore we have

45/100 = P(XY + Z =0) = P(X=0)*P(Y=0)*P(Z=0) + P(X=0)*P(Y=1)*P(Z=0) + P(X=1)*P(Y=0)*P(Z=0)

=(1-a)*(1-b)*(3/4) + (1-a)*b*(3/4) + a*(1-b) *(3/4).

So we have

(1-a)*(1-b) + (1-a)*b + a*(1-b) = 3/5.

then by simple calculation we have 1- a*b = 3/5

that is a*b= 2/5 = 0.4.

Among choices only 0.5 and 0.8 can be the product 0.4.

And the condition a < b says a=0.5 and b=0.8.

Hope it helps.

Best regards,

Jin
Math Revolution