Question

Given: P(A1)= 0.17 P(A2)= 0.07 P(A3)= 0.13 P(A1 OR A2)=0.18 P(A2 OR A3)=0.18 P(A1 OR A3)=0.19 P(A1 AND A2 AND A3)=0.01 Let x be the number of defects that are present.

P(X = 0) = 1 - 0.19 = 0.81

P(A1 OR A2 OR A3) = (0.37) - (0.19) + 0.01 = 0.19 (is this correct ?)

if yes, then is it okay to calculate P(X=2) by summing the intersections P(A1 AND A2) + P(A2 AND A3) + P(A1 AND A3) and subtracting P(A1 AND A2 AND A3) ?

Answer 1

Answer: Yes, your calculation of P(A1 OR A2 OR A3) is correct.

Step-by-step explanation: Your calculation of P(A1 OR A2 OR A3) is correct.

To calculate P(X=2), you can use the formula:

P(X=2) = P(A1 AND A2 AND NOT A3) + P(A1 AND A3 AND NOT A2) + P(A2 AND A3 AND NOT A1)

This is because to have precisely two defects, we need two events A1, A2, and A3 to occur and one not to occur.

You cannot simply sum the intersections P(A1 AND A2) + P(A2 AND A3) + P(A1 AND A3) and subtract P(A1 AND A2 AND A3) to get P(X=2), because this will overcount the cases where all three defects occur together.

Instead, you can use the formula above, and note that:

P(A1 AND A2 AND NOT A3) = P(A1 AND A2) - P(A1 AND A2 AND A3)

P(A1 AND A3 AND NOT A2) = P(A1 AND A3) - P(A1 AND A2 AND A3)

P(A2 AND A3 AND NOT A1) = P(A2 AND A3) - P(A1 AND A2 AND A3)

Substituting these into the formula, we get:

P(X=2) = (P(A1 AND A2) - P(A1 AND A2 AND A3)) + (P(A1 AND A3) - P(A1 AND A2 AND A3)) + (P(A2 AND A3) - P(A1 AND A2 AND A3))

which gives us the correct answer for P(X=2).