# What is the probability of failure?

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

• 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! – Caleb Bouke Oct 14 '17 at 6:14

## 3 Answers

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).

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*}

• Oh thank you so much!! – Caleb Bouke Oct 14 '17 at 6:23

Given: $$P(T\ fails)=0.6, P(B \ fails)=0.3.$$ Then: $$P(T\ works)=0.4, P(B \ works)=0.7.$$ Hence: $$P(exactly \ one \ failure)=1-P(both \ fail)-P(both \ work)=$$ $$1-P(T \ fails)P(B \ fails)-P(T \ works)P(B \ works)=$$ $$1-0.6\cdot 0.3-0.4\cdot 0.7=0.54.$$