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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?

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    $\begingroup$ In this case, loosely, $\text{acceptable} \cap \text{non defective} = \text{acceptable}$. $\endgroup$ – copper.hat Feb 21 '16 at 22:15
  • $\begingroup$ Oh I see, just to verify if I get it, so if $\text{acceptable}$ and $ \text{non defective}$ had no intersection, we would not simplify and if $\text{acceptable}$ contained $ \text{non acceptable}$ then we would eliminate $\text{acceptable}$ ? $\endgroup$ – drossy11 Feb 21 '16 at 22:27
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    $\begingroup$ You lost me there. The point is that if it is acceptable then it is non defective. One set is contained in another, so the intersection is just the acceptable set. $\endgroup$ – copper.hat Feb 21 '16 at 22:28
  • $\begingroup$ Oh okay I get it now. Thank you! $\endgroup$ – drossy11 Feb 21 '16 at 22:29
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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?

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  • $\begingroup$ Oh okay that makes more sense, thank you for the correction. We had the formula I gave in our textbook but the union makes it easier to understand. $\endgroup$ – drossy11 Feb 21 '16 at 22:44

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