Question

Hunters $\mathrm{A}$ and $\mathrm{B}$ shoot at a target; the probabilities of hitting the target are $p_1$ and $p_2$, respectively. Assuming independence, can $p_1$ and $p_2$ be selected so that $P($ zero hits $)=P($ one hit $)=P($ two hits $)?$

Short answer

No, there are no values of $p_1$ and $p_2$ that can be selected such that the probabilities of zero hits, one hit and two hits are equal.

Step-by-step solution

Writing Down the Probability Equations

The probabilities of zero, one and two hits are calculated as follows - The probability of zero hit means neither A nor B hits the target. This would be given by $(1-p_1)\ast (1-p_2)$. The probability of one hit means either A or B hits but not both. This would be given by $(1-p_1)\ast p_2+p_1\ast (1-p_2)$. The probability of two hits is when both A and B hit the target. This would be given by $p_1\ast p_2$. Therefore, our equations are $(1-p_1)\ast (1-p_2)=(1-p_1)\ast p_2+p_1\ast (1-p_2)=p_1\ast p_2$

Solving the Equations

Solving these three equations together may seem complicated. However, since they're all equal to each other, we can set them equal to a common variable, say x. Therefore, we have the system of equations: $(1-p_1)\ast (1-p_2)=x,(1-p_1)\ast p_2+p_1\ast (1-p_2)=x,p_1\ast p_2=x$, up to which we can easily see that $x=0$. If we substitute this into the equations we get $p_1=0$ or $p_2=0$ or $p_1=p_2=1$

Checking the Solutions

Plug these solutions into our system of equations again to check if any of them work. The first solution $p_1=0$ and $p_2=0$ gives us the probabilities of 1 for zero hits, 0 for one hit and 0 for two hits, so it doesn't work. The second solution of $p_1=p_2=1$ gives probabilities of 0 for zero hits, 0 for one hit, and 1 for two hits, so it also doesn't work. Hence, there are no values for p1 and p2 that satisfy the conditions of the problem.

Key concepts

Independent Events

In probability theory, understanding independent events is crucial for analyzing how separate occurrences can affect overall outcomes.
When two events are independent, it means that the outcome of one event does not affect the outcome of the other. In mathematical terms, two events, event A and event B, are said to be independent if the probability of both events happening simultaneously is equal to the product of their individual probabilities. This can be expressed as:

• $P(A\text{ and }B)=P(A)\cdot P(B)$

In the context of the exercise with hunters A and B, their shots at the target are considered independent. That means whether or not hunter A hits the target does not affect the probability of hunter B hitting the target.
Therefore, to analyze the problem's solution, each resulting situation (zero, one, or two hits) can be calculated individually using their independent probabilities.

Probability Equations

Probability equations help us express the likelihood of different outcomes in a quantifiable manner. In this particular exercise, we start by defining the events:

• The chance of zero hits, when neither A nor B hits the target, is derived from the probabilities of missing the target for both: $(1-p_1)\cdot (1-p_2)$.
• The chance of one hit involves two scenarios, either hunter A hits while B misses, or B hits while A misses, which gives us: $(1-p_1)\cdot p_2+p_1\cdot (1-p_2)$.
• Lastly, the probability of both hitting the target (two hits) is simply: $p_1\cdot p_2$.

These equations form the base for further calculations by setting them equal to determine possible values of $p_1$ and $p_2$ that satisfy equality among all outcomes.
Such equations allow us to answer more complex questions by equating different scenarios and solvable expressions.

System of Equations

When dealing with multiple probability equations at once, combining them into a system of equations can streamline the solution process. A system of equations allows mathematicians to solve multiple equations simultaneously for common variables.
In the exercise provided, we must solve a system involving three equations based on different probability outcomes:

• $(1-p_1)\cdot (1-p_2)=x$
• $(1-p_1)\cdot p_2+p_1\cdot (1-p_2)=x$
• $p_1\cdot p_2=x$

By setting these equations equal to a common variable $x$, we examine the conditions under which each equation balances equally with the others.
Throughout the exercise, solving these equations shows if particular values for $p_1$ and $p_2$ allow all three scenarios to have identical probabilities.
Ultimately, the conclusion is reached by checking the conditions under which these probabilities equalize, thereby validating if such $p_1$ and $p_2$ values exist.