I have tried to solve a conditional probability problem but struggle to get it right.

At a workplace 1% of the staff where injured during a year. 60% of all injured where men. 30% of the employees were women. Is it male or female employees that has the biggest risk of getting injured?

The answer says women.

This is how I have tried to solve the problem.

Let A be the event of men, B be the event of women and I the event of injured.

  • P(A): 0.7
  • P(B): 0.3
  • P(I|A): 0.6
  • P(I|B): 0.4

$$ P(I) = P(I \cap A) + P(I \cap B) = P(A)P(I \mid A)+ P(B)P(I \mid B) \Rightarrow P(I) = 0.7*0.6 + 0.3*0.4 $$

Then I try to solve the probability for men given they have been injured:

$$ P(A \mid I) = \frac{P(A \cap I)}{P(I)} \Leftrightarrow \frac{P(A)P(I \mid A)}{P(I)} \Rightarrow \frac{0.7*0.6}{0.7*0.6+0.3*0.4} = 0.78 $$

After getting this answer there seems to be some problem. I think that I don´t take into account that the total of injured is 1% of the employees. Should I just use the fact that P(I) is equal to 0.01? I think I need some hint to be able to solve this problem.

  • $\begingroup$ $P(I|A)=0.6$ suggests that working there gives me a chance of $60$ percent to be injured... No, I will not get to work there. $\endgroup$
    – drhab
    Oct 20 '15 at 9:56

Your mistake is in your first probabilities. "60% of injuries are male" can be re-formulated: "the probability that an injury is a male is 0.6" ie $P(A|I)=0.6$ using your notations. You said the opposite: $P(I|A)=0.6$, which means that 60% of males are injured (by the way, this is exactly what you want to compute)!

To solve your problem, you just have to use Bayes rule:

$P(I|A) = \frac{P(A|I) P(I)}{P(A)} = \frac{P(A|I) P(I)}{1 - P(B)} = \frac{0.6 * 0.01}{0.7}=0.0086$


$P(I|B) = \frac{P(B|I) P(I)}{P(B)} = \frac{0.4 * 0.01}{0.3}=0.013$

The answer is women, as you were told.


The problem lies with $P(I \mid A) = 0.6$ and $P(I \mid B) = 0.4$. They're the wrong way around. We know that 60% of the injured are men, or in other words, if someone is injured, it's a man with 60% probability. So $P(A \mid I) = 0.6$ and $P(B \mid I) = 0.4$.

What you want to know is whether $P(I \mid A) \gt P(I \mid B)$. Use Bayes' rule as before:

$P(I \mid A) = \frac{P(A \mid I) P(I)}{P(A)} = \frac{0.6 \times 0.01}{0.7} \approx 0.8\%$

$P(I \mid B) = \frac{P(B \mid I) P(I)}{P(B)} = \frac{0.4 \times 0.01}{0.3} \approx 1.3\%$

So a person has greater probability of being injured if they're female than if they're male.


Not really an answer, but it can make things less abstract.

Let it be that $1000$ persons are working at the workplace. Then $10$ persons were injured and $6$ of them were men. Consequently $4$ women were injured. There are $300$ women and $1000-300=700$ men working there.

Probability for a man to be injured: $\frac{6}{700}$.

Probability for a woman to be injured: $\frac{4}{300}$.


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