# Difference in Prior and Posterior Probability

I'm a bit puzzled by the answer I'm getting in a quite simple problem.

$$4$$ athletes simultaneously shoot at the target. It is known that the probability of hitting the first hunter is $$0.4$$, second - $$0.6$$, third - $$0.7$$, fourth - $$0.8$$. As a result, one of the athletes did not hit the target. What is the probability that it was the second athlete?

I'll give my solution:

Event $$A$$ - the second athlete did not hit the target

Event $$B$$ - $$3$$ out of $$4$$ athletes hit the target

Based on the condition of the problem, I thought that we need to find $$P(A|B)$$.

So, $$P(A|B)=\frac{P(B|A) \cdot P(A)}{P(B)} =\frac{(0.4 \cdot 0.4 \cdot 0.7 \cdot 0.8) \cdot 0.4}{0.4 \cdot 0.6 \cdot 0.7 \cdot 0.2 + 0.4 \cdot 0.6 \cdot 0.3 \cdot 0.8 + 0.4 \cdot 0.4 \cdot 0.7 \cdot 0.8 + 0.6 \cdot 0.6 \cdot 0.7 \cdot 0.8 } = 0.0937$$

It seems that the found probability turned out to be extremely small. Have I correctly defined what is asked to be found in this problem in terms of elementary events?

It also confuses me that the sum of the posterior probabilities of not hitting the target of each athlete comes out to be significantly less than one. A posteriori knowledge that only one of the shooters did not hit the target, in my opinion, should greatly increase the probability of not hitting for the "weakest" shooters - the first and second. But the two of them together will account for about 40%.

• The problem isn't clear. Did you mean to assume that exactly one athlete missed or that at least one athlete missed?
– lulu
May 1, 2022 at 20:19
• Exactly one athlete May 1, 2022 at 20:39

Let's look at two cases depending on whether the given condition is that exactly one athlete missed, versus at least one athlete missed.

Case 1: Exactly one athlete missed. Then according to your event definitions, we have

$$\Pr[A \mid B] = \frac{\Pr[B \mid A]\Pr[A]}{\Pr[B]}.$$

Then $$\Pr[B \mid A]$$, the probability that exactly one athlete missed given that the second athlete missed, is equivalent to the probability that the first, third and fourth athlete hits: $$\Pr[B \mid A] = (0.4)(0.7)(0.8) = 0.224.$$ The marginal probability that exactly one athlete missed is \begin{align} \Pr[B] &= (1 - 0.4)(0.6)(0.7)(0.8) + (0.4)(1 - 0.6)(0.7)(0.8) \\ &\quad+ (0.4)(0.6)(1 - 0.7)(0.8) + (0.4)(0.6)(0.7)(1 - 0.8) \\ &= 0.3824. \end{align} Therefore, the desired posterior probability is $$\Pr[A \mid B] = \frac{(0.224)(0.4)}{0.3824} = 0.23431.$$

You have an extra factor of $$0.4$$ in the numerator. I suspect this is because you mistakenly computed $$\Pr[B \cap A]\Pr[A]$$ instead of $$\Pr[B \mid A]\Pr[A]$$.

Case 2: At least one athlete missed; call this event $$B'$$. Then $$\Pr[B' \mid A] = 1$$, since given that the second missed, at least one missed.

The computation of the denominator is also simplified, since the unconditional probability that at least one athlete missed is the complement of the probability that no athlete missed, which is simply $$(0.4)(0.6)(0.7)(0.8) = 0.1344$$. Therefore, the desired posterior probability is $$\Pr[B'] = 1 - 0.1344 = 0.8656$$ and $$\Pr[A \mid B'] = \frac{\Pr[B' \mid A]\Pr[A]}{\Pr[B']} = \frac{0.4}{0.8656} = 0.462107.$$

$$4$$ athletes simultaneously shoot at the target. It is known that the probability of hitting the first hunter is $$0.4$$, second - $$0.6$$, third - $$0.7$$, fourth - $$0.8$$. As a result, one of the athletes did not hit the target. What is the probability that it was the second athlete?

Alternative approach:

My answer corresponds to Case 1 of heropup's answer, which reflects the comment given by the original poster after heropup posted his answer: the original poster intended that exactly one shooter missed.

heropup's answer (in effect) diagnoses where the original poster went wrong. So, it is open season to present an alternative approach.

First of all, it is assumed that each shooter's probability of hitting the target is normally independent of whether any of the other shooters hit the target. The problem can not be attacked without such an assumption.

Event $$B$$ is the event that exactly one of the four missed.
Event $$A$$ is the event that the 2nd shooter missed.

The question asked for $$p(A|B)$$.

You know that $$p(B) \times p(A|B) = p(A ~\text{and}~ B)$$.

Therefore, the desired computation is

$$\frac{p(A ~\text{and}~ B)}{p(B)}. \tag1$$

To me, it seems simplest to assign variables, derive the formula in terms of these variables, and then evaluate the formula numerically.

$$\text{Let}~~ (p_1,p_2,p_3,p_4) ~~\text{denote}~~ (0.4,0.6,0.7,0.8). \tag2$$

$$\text{Let}~~ (q_1,q_2,q_3,q_4) ~~\text{denote}~~ (1-p_1,1-p_2,1-p_3,1-p_4). \tag3$$

Then, in (1) above, the denominator is given by

$$(p_1p_2p_3q_4) + (p_1p_2q_3p_4) + (p_1q_2p_3p_4) + (q_1p_2p_3p_4). \tag4$$

In (1) above, the numerator is given by:

$$(p_1q_2p_3p_4). \tag5$$

So, it only remains to substitute the numbers from (2) and (3) into (4) and (5) above, and then construct the corresponding numerical fraction that represents (1) above.

The evaluation of (4) above is

$$(0.4 \times 0.6 \times 0.7 \times 0.2) + (0.4 \times 0.6 \times 0.3 \times 0.8)$$

$$+ (0.4 \times 0.4 \times 0.7 \times 0.8) + (0.6 \times 0.6 \times 0.7 \times 0.8)$$

$$=~ \frac{1}{10000} \times [336 + 576 + 896 + 2016] = \frac{3824}{10000}. \tag6$$

The evaluation of (5) above is

$$\frac{896}{10000}. \tag7$$

Therefore, using (6) and (7) above, the expression in (1) above evaluates to

$$\frac{896}{3824} = \frac{56}{239} \approx (0.234).$$