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There are three probable events.

A = "clear weather" = $0.60$, B = "rainy" = $0.39 $, C = "snowy" = $0.01 $

The probability of coming in a accident on one of theese conditions are respectably

$0.001$ on clear weather, $0.003$ on rainy and $0.009$ on a snowy day.

What is the probability of coming in a accident?

Here i just multiply each of the conditional probabilities, and come up with a answer of $0.0019$.

$$P(A) \times 0.001 + P(B) \times 0.003 + P(C) \times 0.009 = 0.0019 $$

Now that answer is correct, but the second one is really hard to understand. It goes as follows. Given that you get in an accident, what are the probability that snowy weather was the reason?

  • 0
    Create a probability tree, then such questions are easy to handle. You will have to use Bayes law for the last equation, which is easy to read out from the probability tree. – 2017-02-28
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    Heuristically, you can think of this second question as follows: You obtained your probability of $0.0019$ (actually $0.00186$) by adding up three joint probabilities, each one corresponding to one of sunny, rainy, or snowy weather. What fraction of that $0.00186$ was due to the contribution from snowy weather? In this case, it is $0.00009/0.00186 = 3/62$. – 2017-02-28

3 Answers 3

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Define events $C$, $R$, $S$ to mean that the weather is clear, rainy, and snowy, respectively. Define event $A$ to mean that an accident is observed.

With this notation, your desired probability for the second part of your question is $\Pr[S \mid A]$, the probability that the weather was snowy if an accident was observed.

Bayes' theorem then gives $$\Pr[S \mid A] = \frac{\Pr[A \mid S]\Pr[S]}{\Pr[A]}.$$ You have already calculated the denominator in the first part: $$\Pr[A] = \Pr[C]\Pr[A \mid C] + \Pr[R]\Pr[A \mid R] + \Pr[S]\Pr[A \mid S] \approx 0.00186.$$ (You have rounded this to $0.0019$ but I advise to keep as many significant figures as you can for the next step.) All that is left now is to calculate the numerator. But you have also done this part too; it is simply the third term in the sum above: $$\Pr[A \mid S] \Pr[S] = (0.009)(0.01).$$ So, your desired probability is $$\Pr[S \mid A] = \frac{(0.009)(0.01)}{0.00186} \approx 0.0483871.$$

Some helpful tips:

  • Choose good notation; i.e., choose labels that relate to the events of interest.
  • Questions that have multiple parts usually (but not always) are designed so that the answers to previous parts will relate to the solutions for following parts.
  • Use formulas and relationships that you have learned.
  • Ask yourself if your answer is consistent with your intuition. How can you test the reasonableness of your answer?
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$P(A) = .6$, $P(B) = .39$, $P(C) = .01$ Now, let $P(D)$ = Probability of an accident.

We know that: $P(D|A) =.001$, $P(D|B) = .003$, and $P(D|C) = .009$

What we want to know is $P(C|D)$ so we use Bayes theorem.

$P(C|D) = {{P(D|C)P(C)}\over{P(D|C)P(C)+P(D|A)P(A)+P(D|B)P(B)}}$

So plug in what we know now

$(.009)(.01)\over{(.009)(.01)+(.001)(.6)+(.003)(.39)}$

= $.0483870968$

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Sometimes Bayes Law can be very overwhelming with the equation. I like to visualize it for simplicity. So allow me to throw my diagram and to look at your question from a different angle (with the same result):

       clear 
       weather (0.6)
      /(A)-------------- 0.001---------> accident = 0.0006
     /
    /
   /
  /    rainy (0.39)
  -----(B)--------------- 0.003--------> accident = 0.00117 
  \
   \
    \
     \ 
      \ 
       \(C) (0.01)
        snowy------------- 0.009 -------> accident = 0.00009

so you have a probability of: 

        0.00009
---------------------- = 0.0483
0.00009+0.00117+0.0006