Question

$A$and $B$are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of $A$will hit $B$with probability $pA$, and each shot of $B$will hit $A$with probability $pB$. What is

(a) the probability that $A$is not hit?

(b) the probability that both duelists are hit? (c) the probability that the duel ends after the nth round of shots

(d) the conditional probability that the duel ends after the nth round of shots given that $A$is not hit?

(e) the conditional probability that the duel ends after the nth round of shots given that both duelists are hit?

Short answer

1. The required probability for part (a) is,$\frac{p_A\left(1-p_B\right)}{p_A+p_B-p_A{p}_B}$
2. The required probability for part (b) is,$\frac{p_A{p}_B}{p_A+p_B-p_A{p}_B}$
3. The required probability for part (c) is,${\left[\left(1-p_A\right)\left(1-p_B\right)\right]}^{n-1}\left[p_A+p_B-p_A{p}_B\right]$
4. The required probability for part (d) is,$\left.{\left(1-p_A\right)}^{n-1}{\left(1-p_B\right)}^{n-1}\left[p_A+p_B-p_A{p}_B\right]\right]$
5. The required probability for part (e) is,$\left.{\left(1-p_A\right)}^{n-1}{\left(1-p_B\right)}^{n-1}\left[p_A+p_B-p_A{p}_B\right]\right]$
   

Step-by-step solution

Given information(part a)

$A$and $B$are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of A will hit B with probability$P_A$, and each shot of localid="1646911603210" $B$will hit localid="1646911609293" $A$with probability localid="1646911618881" $p_B$.

Step 2:Explanation(part a)

According to the question, there are two events $A$and $B$who are involved in a duel. Moreover, the probability that each shot of $A$will hit $B$with probability $p_A$and the probability that each shot of $B$will hit $A$with probability $p_B$.

Calculation (part a)

According to the question, the duel ends if one or both are hit. That is, the duel ends if at least one of them hits.

The probability that $A$is not hit given that the duel ends can be computed is follows:

$P(A$is not hit $\mid$Duelends$)=P(A$ is not hit $\mid$atleast one of $A$and $B$hits $)$

localid="1646911755249" $=\frac{P\left(A\text{is not hit and atleast one hits}\right)}{P\left(\text{atleast one of}A\text{and}B\text{hits}\right)}$

localid="1646911760035" $=\frac{P\left(A\text{hits}B\right)P\left(B\text{does not hit}A\right)}{P\left(\text{atleast one of}A\text{and}\mathrm{B}\text{hits}\right)}$

The probability that at least one of localid="1646911765885" $A$and localid="1646911770963" $B$hits is calculated as follows:

localid="1646911838565" $P($atleast one of localid="1646911848513" $A$and localid="1646911855691" $\mathrm{B}$hits localid="1646911776135" $)=1-P(A$is not hit and localid="1646911865394" $B$is not hit $)$

localid="1646911872845" $=1-(1-P($A is hitlocalid="1646911879089" $\left)\right)\times (1-P(B$is hitlocalid="1646911887606" $\left)\right)$

localid="1646911893697" $=1-\left(1-p_B\right)\left(1-p_A\right)$

localid="1646911899914" $=p_A+p_B-p_A{p}_B$

The required Probability (part a)

The required probability is calculated as follows,

$P(A$is not hit $\mid$Duel ends $)=\frac{\frac{P\left(A\text{hits}B\right)P\left(B\text{does not hit}A\right)}{P\left(\text{atleast one of}A\text{and}\mathrm{B}\text{hits}\right)}}{P\left(\text{atleast one of}A\text{and}\mathrm{B}\text{hits}\right)}$

$=\frac{p_A\left(1-p_B\right)}{p_A+p_B-p_A{p}_B}$

Hence,

the probability that $A$is not hit is

$\frac{p_A\left(1-p_B\right)}{p_A+p_B-p_A{p}_B}.$

Step 5:Final answer(part a)

The probability that $A$is not hit is

$\frac{p_A\left(1-p_B\right)}{p_A+p_B-p_A{p}_B}.$

Given information (part b)

$A$and $B$are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of $A$will hit $B$with probability $pA$, and each shot of $B$will hit $A$with probability $pB$.

Step 7:Explanation(part b)

According to the question, there are two events $A$and $B$who are involved in a duel. Moreover, the probability that each shot of $A$will hit $B$with probability $p_A$and the probability that each shot of$B$will hit $A$with probability $p_B$.

The probability that both duelists are hit  (part b)

The duel ends only if one or both are hit, that is, the duel ends if at least one of them is hit The probability that both duelists are hit given that the duel ends can be computed as follows:

$P($Both duelists are hit $\mid$Duel ends$)=P($Both duelists are hit $\mid$atleast one of $A$and $B$is hit $)$

localid="1646912187403" $=\frac{P\left(\begin{array}{l}\text{Both duelists are hit and}\\ \text{atleast one of}A\text{and}B\text{is hit}\end{array}\right)}{P\left(\text{atleast one of}A\text{and}B\text{is hit}\right)}$

localid="1646912194580" $=\frac{P\left(\text{both duelistsare hit}\right)}{P\left(\text{at least one of}A\text{and}B\text{is hit}\right)}$

The probability that both duelists are hit is calculated as follows:

localid="1646912207490" $P($both duelists are hit localid="1646912200888" $)=P(A$hitslocalid="1646912213720" $B$and localid="1646912220027" $B$hits localid="1646912226642" $A)$

localid="1646912233426" $=P(A$hits localid="1646912239275" $B)\times P(B$hits localid="1646912252407" $A)$

localid="1646912262959" $=p_A{p}_B$

The required probability (part b)

The required probability is calculated as follows:

$P(\text{Both duelists are hit}\mid \text{Duel ends})=\frac{P\left(\text{both duelists are hit}\right)}{P\left(\text{at least one of}A\text{and}B\text{is hit}\right)}$

$=\frac{p_A{p}_B}{p_A+p_B-p_A{p}_B}$

Hence, the required probability is

$\frac{p_A{p}_B}{p_A+p_B-p_A{p}_B}$

Final answer(part b)

The required probability is

$\frac{p_A{p}_B}{p_A+p_B-p_A{p}_B}.$

Step 11. Given information(part c)

$A$and $B$are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of$A$will hit $B$with probability $pA$, and each shot of $B$will hit $A$with probability $pB.$

Step 12:Explanation(part c)

According to the question, there are two events $A$ and $B$ who are involved in a duel. Moreover, the probability that each shot of $A$ will hit $B$ with probability $p_A$ and the probability that each shot of $B$ will hit $A$ with probability $P_B$.

Step 13:Explanation(part c)

Probability that the duel ends after the $n^{th}$round of shots is equal to the probability that there are no hits for the first $(n-1)$rounds and there is at least one hit on the $n^{\text{th}}$round.

Therefore, the required probability is calculated as follows:

$P\left(\begin{array}{l}\text{no hits}(n-1)\text{rounds}n\\ \text{at least one hit}{n}^{\text{th}}\text{round}\end{array}\right)=P\left[\begin{array}{l}\left(\begin{array}{l}A\text{does not hit}B\text{and}B\text{does}\\ \text{not hit}A\text{in}(n-1)\text{rounds}\end{array}\right)\\ \cap \left(\begin{array}{l}\text{atleast one of}A\text{or}B\\ \text{is hitat}{n}^{\text{th}}\text{round}\end{array}\right)\end{array}\right]$

$=\left(\begin{array}{l}P\left(A\text{doesnot hit}B\text{in}\right(n-1\left)\text{rounds}\right)\\ \times P\left(B\text{does not hit}A\text{in}\right(n-1\left)\text{rounds}\right)\\ \times P\left(\text{atleast one of}A\text{or}B\text{is hitat}{n}^{\text{th}}\text{round}\right)\end{array}\right)$

$={\left(1-p_A\right)}^{n-1}{\left(1-p_B\right)}^{n-1}\left[p_A+p_B-p_A{p}_B\right]$

$={\left[\left(1-p_A\right)\left(1-p_B\right)\right]}^{n-1}\left[p_A+p_B-p_A{p}_B\right]$

Hence, the required probability is

${\left[\left(1-p_A\right)\left(1-p_B\right)\right]}^{n-1}\left[p_A+p_B-p_A{p}_B\right]$

Step 14:Final answer(part c)

The required probability is

${\left[\left(1-p_A\right)\left(1-p_B\right)\right]}^{n-1}\left[p_A+p_B-p_A{p}_B\right]$

Given information(part d)

$A$and$B$are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of $A$will hit $B$with probability $pA$, and each shot of$B$will hit $A$with probability$pB$.

Explanation(part d)

According to the question, there are two events $A$and $B$who are involved in a duel. Moreover, the probability that each shot of $A$will hit $B$with probability $p_A$and the probability that each shot of $B$will hit $A$with probability $p_B$.

Step 17:The conditional Probability (part d)

The conditional probability that the duel ends after the nth round of shots given that $A$is not hit is calculated as follows:

$P\left(\begin{array}{l}\left.\text{Duelendsafter}{n}^{\text{th}}\text{round}\mid \right)=\frac{P\left(\text{Duelendsafter}{n}^{\text{th}}\text{round and}A\text{is not hit}\right)}{P\left(A\text{is not hit}\right)}\end{array}\right.$

$=\frac{P\left(\begin{array}{l}\text{Duelendsafter}{n}^{\text{th}}\text{round and}\\ \text{Ais not hit,}B\text{is hit}\end{array}\right)}{P\left(A\text{is not hit}\right)}$

$=\frac{{\left(1-p_A\right)}^{n-1}{\left(1-p_B\right)}^{n-1}\left(1-p_B\right)\left(p_A\right)}{\frac{p_A\left(1-p_B\right)}{p_A+p_B-p_A{p}_B}}$

$={\left(1-p_A\right)}^{n-1}{\left(1-p_B\right)}^{n-1}\left[p_A+p_B-p_A{p}_B\right]$

Hence, the required probability is

${\left(1-p_A\right)}^{n-1}{\left(1-p_B\right)}^{n-1}\left[p_A+p_B-p_A{p}_B\right]$

Step 18:Final answer(part d)

The required probability is

${\left(1-p_A\right)}^{n-1}{\left(1-p_B\right)}^{n-1}\left[p_A+p_B-p_A{p}_B\right]$

Given information(part e)

$A$and $B$are involved in a duel. The rules of the duel are that they are to pick up their guns and shoot at each other simultaneously. If one or both are hit, then the duel is over. If both shots miss, then they repeat the process. Suppose that the results of the shots are independent and that each shot of $A$will hit $B$with probability $pA$, and each shot of B will hit A with probability $pB$.

Step 20:Explanation(part e)

According to the question, there are two events $A$and $B$who are involved in a duel. Moreover, the probability that each shot of $A$will hit $B$with

probability $p_A$and the probability that each shot of $B$will hit $A$with probability $p_B$.

The conditional probability (part e)

The conditional probability that the duel ends after the $nth$round of shots given that both duelists are hit is calculated as follows:

$P\left(\begin{array}{l}\text{Duelendsafter}{n}^{\text{W}}\text{round}\mid \\ \text{Both duelistsarehit}\end{array}\right)=\frac{\mathrm{P}\left(\text{duel ends after}n\text{th round and both duelists are hit}\right)}{\mathrm{P}\left(\text{both duelists are hit}\right)}$

$=\frac{{\left(1-p_A\right)}^{n-1}{\left(1-p_B\right)}^{n-1}\left(p_B\right)\left(p_A\right)}{\frac{p_B{p}_A}{p_A+p_B-p_A{p}_B}}$

$={\left(1-p_A\right)}^{n-1}{\left(1-p_B\right)}^{n-1}\left[p_A+p_n-p_A{p}_B\right]$

Hence, the required probability is

${\left(1-p_A\right)}^{n-1}{\left(1-p_B\right)}^{n-1}\left[p_A+p_B-p_A{p}_B\right]$

Final answer(part e)

The required probability is

${\left(1-p_A\right)}^{n-1}{\left(1-p_B\right)}^{n-1}\left[p_A+p_B-p_A{p}_B\right]$