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I had an AP Statistics final exam today and there was one multiple choice question which stumped me. It was along the lines of the following:

  • Two independent traffic lights
  • The probability that both are red is $0.22$
  • The probability that the first is red and the second is not is $0.33$
  • What is the probability that the second traffic light is red?

Options:

A) $0.40$

B) $0.45$

C) $0.50$

D) $0.55$

E) $0.60$

I was told that the correct answer is $0.40$ but I couldn't figure out how to get there. Any tips? Thanks.

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3 Answers 3

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let $p$ and $q$ be the probability that the first and second lights are red respectively.

Since we have independence we are essentially told:

$pq=0.22$ and $p(1-q)=0.33$

dividing both equation we deduce: $\frac{q}{1-q}=\frac{2}{3}$.

this equation is solved readily: $\frac{q}{1-q}=\frac{2}{3}\iff 3q=2-2q\iff q=\frac{2}{5}$.

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So $pq= 0.22$ and $p(1-q) = 0.33$. Consider $p(1-q)$ as $p - pq$ (foil it / distributive property). Plug in $0.22$ to $pq$, so $p-0.22= 0.33$, $p = 0.55$. Now plugging in again, $pq =0.22$, $0.55q=0.22$, $q=0.22/0.55= 0.40$.

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  • $\begingroup$ See the introduction to posting mathematical notation. While it is helpful to have symbols $p,q$ to represent some probabilities in the problem, it is required to state (in words) what these are meant to represent. They are not used in the Original Post, so it is your responsibility to identify what these represent. $\endgroup$
    – hardmath
    Commented May 12, 2018 at 1:41
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Note: $$P(F\cap S)+P(F\cap S')=P(F)=0.55;\\ P(F\cap S)=P(F)\cdot P(S)=0.22; \\ P(S)=\frac{0.22}{0.55}=\frac25.$$

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