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The Lemon Brand cars have a transmission which fails with probability $0.6$, and brakes which fail with probability $0.3$; the two kinds of failures occur independently. What is the probability that exactly one of the failures occurs when you drive a Lemon?

I'm very confused with the question because I tried doing the Bernoulli Trials but I don't have (n) or Probability of success since there isn't a number of trials that I can base it off of. I did just try and multiply together to get $0.18$ but it is not the answer. Could any body give me some sort of a hint?

*The answer is .54

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  • $\begingroup$ I know the answer because there is an answer sheet but I really want to know how to get that number so I can study for the test! $\endgroup$ – Caleb Bouke Oct 14 '17 at 6:14
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Hint. The (probability that exactly ONE of the failures occurs) is equal to (the probability that the transmission fails and brakes work) plus (the probability that transmission works and brakes fail).

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Say there are $100$ such cars. Of these, $60$ have transmissions that will fail, $30$ will have brakes that will fail. Because the types of failure are independent, this also means that $18 = 100(0.6)(0.3)$ of these cars will fail in both ways. This means that there are $60 + 30 - 2(18)$ cars that fail in one way or the other but not both.

In a more formal notation, let $T$ be the event that a randomly selected car will have a transmission failure, and let $B$ be the event that a randomly selected car will have a brake failure. Then $$\Pr[T] = 0.6, \quad \Pr[B] = 0.3.$$ By the definition of independence, we find $$\Pr[T \cap B] = \Pr[T]\Pr[B] = (0.6)(0.3) = 0.18.$$ It follows that $$\begin{align*} \Pr[T \cap \bar B] + \Pr[\bar T \cap B] &= \Pr[T]\Pr[\bar B] + \Pr[\bar T]\Pr[B] \\ &= \Pr[T] (1 - \Pr[B]) + (1 - \Pr[T])\Pr[B] \\ &= (0.6)(1 - 0.3) + (1 - 0.6)(0.3). \end{align*}$$

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  • $\begingroup$ Oh thank you so much!! $\endgroup$ – Caleb Bouke Oct 14 '17 at 6:23
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Given: $$P(T\ fails)=0.6, P(B \ fails)=0.3.$$ Then: $$P(T\ works)=0.4, P(B \ works)=0.7.$$ Hence: $$P(exactly \ one \ failure)=1-P(both \ fail)-P(both \ work)=$$ $$1-P(T \ fails)P(B \ fails)-P(T \ works)P(B \ works)=$$ $$1-0.6\cdot 0.3-0.4\cdot 0.7=0.54.$$

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