Formulate and solve task in terms of probabilities

Question

I want to formulate task in terms of probabilities and solve it.

Let's say $H$ - helmet, $A$ - accident, $I$ - injury. I know that probability of an accident with a helmet is 40% higher than without a helmet, so it's $P(A|H) = (1+0.4) * P(A | \overline{H})=1.4* P(A | \overline{H})$. I also know that a helmet decrease number of injuries by 40%, so $P(I|A, H) = (1-0.4)* P(I|A, \overline{H})=0.6*P(I|A, \overline{H})$. I want to find ratio of probabilities of injuries with helmet and without it, $\frac{P(I|H)}{P(I|\overline{H})}=?$

Summary:

We have

$P(A|H) =1.4* P(A | \overline{H})$

$P(I|A, H) =0.6*P(I|A, \overline{H})$

Find $\frac{P(I|H)}{P(I|\overline{H})}=?$

I'm not sure if I formulate the task correctly, hope you can help me with this. And Anyone has an idea how to calculate the ratio? Or maybe anyone knows similar task?

Answer 1

As you established, we have:

$$P(A | H) = 1.4 P(A | \bar{H})$$

$$P(I | H, A) = 0.6 P(I | \bar{H}, A)$$

Then, the probability of having an injury given that you are not wearing a helmet, equals:

$$P(I | \bar{H}) = P(I | \bar{H}, A) P(A | \bar{H}) + P(I | \bar{H}, \bar{A}) P(\bar{A} | \bar{H})$$

Basically, you distinguish between two cases: $A$ and $\bar{A}$. Since a person cannot be injured without an accident, the second term equals $0$. Thus, we find:

$$P(I | \bar{H}) = P(I | \bar{H}, A) P(A | \bar{H})$$

Similarly, for the case where a helmet is being worn, you find that:

$$P(I | H) = P(I | H, A) P(A | H) + P(I | H, \bar{A}) P(\bar{A} | H)$$

Again, the second term equals $0$. If we now plug in the first two equations, we find:

$$P(I | H) = P(I | H, A) P(A | H) = 0.6 P(I | \bar{H}, A) 1.4 P(A | \bar{H}) = 0.84 P(I | \bar{H}, A) P(A | \bar{H})$$

Thus, we find:

$$\frac{P(I | H)}{P(I | \bar{H})} = 0.84$$

This essentially means that the probability of an injury will be reduced by $16\%$ when wearing a helmet.

You can also look at it this way: if the number of accidents goes up by $40\%$ and the number of injuries in case of an accident goes down by $40\%$, the relative number of injuries should go down to $1.4 \cdot 0.6 = 0.84$.