Probability of failing a test for A or B are $A=0.5$ and $B=0.2$.

Probability that at least one will fail is $0.5+0.2-(0.5)(0.2)=0.6$

Probability that both passing is $1-0.6=0.4$.

I thought I could get the probability of A and B passing by doing $1-A=0.5$ and $1-B=0.3$ and using the same equation on those numbers? Why isn't the probability of both passing $0.5+0.3-(0.5)(0.3)=0.65$? What's the problem in my logic of subtracting to get the probability of passing?

  • $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos Aug 3 '17 at 7:01
  • $\begingroup$ Please check your math....$1-P(B)\neq 0.3$. Also, remember that $(A \cup B)^c = A^c \cap B^c$ $\endgroup$ – user408433 Aug 4 '17 at 2:13

Assuming that the tests are independent:

$P(\text {A }\land \text{B pass}) = P(\text{A pass}) \times P(\text {B pass})= (1-0.5)(1-0.2) = 0.4 $

The 2nd equality because probabilities add up to 1 and the events are presumably complements.

Note that $1-P(B) = 0.2$, so that:

$P(\text{A fail} \space \lor \space \text {B fail})= P(\text{A fail)}+P(\text{B fail}) - P(\text {A }\land \text{B fail}) = 0.5+0.2-0.5\times0.2 =0.6$

And the complement of that is indeed $P(\text {A }\land \text{B pass})$.

(Sorry for using logic notation).


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