A bin contains 5 defective (that immediately fail when put in use), 10 partially defective (that fail after a couple of hours of use), and 25 acceptable transistors. A transistor is chosen at random from the bin and put to use. If it does not immediately fail, what is the probability it is acceptable?
Intuitively, I got the answer that I am pretty sure is right:
$$ \frac{P(\text{acceptable})}{P(\text{not defective})} = \frac{25/40}{35/40} = 5/7$$
I'm having a hard time simplifying the conditional probability formula to get the same answer.
Probability of $E$ given $F = P(E|F) = \frac{P(EF)}{P(F)} $
So in this problem we have
$$ P(\text{acceptable} | \text{non defective}) = \frac{P(\text{acceptable} \times \text{non defective})}{P(\text{non defective})}$$
$$ = \frac{25/40 \times 35/40}{35/40} = 25/40 = 5/8$$
Which is not the same answer, how do I make my assumption fit with the conditional probability formula?
Best Answer
It is unclear what you meant when you said: Probability of $E$ given $F = P(E|F) = \frac{P(EF)}{P(F)} $
The correct statement is: Probability of $E$ given $F = P(E|F) = \frac{P(E \cap F)}{P(F)} $
It is important to note that in general $P(E \cap F) \ne P(E) \times P(F)$
Then $P(\text{acceptable} | \text{non defective})= \frac{P(\text{acceptable} \cap \text{non defective})}{P(\text{non defective})}$
$ = \frac{25/40}{35/40} = \frac 57 $
Which is not the same answer, how do I make my assumption fit with the conditional probability formula?