Each of the variables X, Y, and Z can
only be 0 or 1. The values of the three variables are independent of one another.
The probability that Z is 1 is
.25. The probability that X is
1 is less than the probability that Y is
1. The probability that XY + Z is at least 1 is .55.

In the table below, select two numbers that are consistent with the
information that is given. In the first column, select the row that shows
the probability that X is 1, and
in the second column, select the row that shows the probability that Y is 1. Make exactly one selection for each
column.

answers are in square box. pls explain how to solve this problem.

Hi

P[ (XY+ Z) >=1 ] = 0.55,  therefore P[ (XY + Z) < 1] = 0.45
P[Z=1] = 0.25 so P[Z=0] = 0.75

Now for (XY + Z) < 1 to be true  XY should be 0  and also Z= 0 .

for XY = 0 there are three possibilities:
Condition1.    X=0 and Y=0
Condition 2.   X=1 and Y=0
Condition 3.   X=0 and Y=1

so P[ (XY=0)] = P(Condition1) + P(Condition 2) + P(Condition 3)
Lets call , P[X=1] = a and P[Y=1] = b , therefore P[X=0]=1-a and P[Y=0] = 1-b

so , 0.45 = P[XY=0] * p[Z=0]
    0.45 = ((1-a) *b  + (1-b)*a + (1-a)*(1-b) ) * 0.75
   0.6 = 1-ab
   finally , ab  = 0.4 , since a < b therefore a=0.5 and b=0.8

Hope it helps 🙂