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?

A and

jeffp

Answered by Aaron 2 years ago

let

$P(AB^c)=p_B(1-p_A)$ be the event that A is hit and B is not hit

$P(A^cB)=(1-p_B)p_A$ be the event that A is not hit and B is hit

$P(AB)=p_Bp_A$ be the event that A is hit and B is hit

$P(A^cB^c)=(1-p_B)(1-p_A)$ be the event that A is hit and B is not hit

**the probability that A is not hit**

we are looking for the probability

$$P(A)=P(A|AB^c)P(AB^c)+P(A|A^cB)P(A^cB)+P(A|AB)P(AB)+P(A|A^cB^c)P(A^cB^c)$$

recognize that if the event $P(A|A^cB^c)$ is the event that they both miss, and so they must start over again meaning $P(A|A^cB^c)=P(A)$ and therefore

$$P(A)=1 \cdot p_B(1-p_A)+0\cdot (1-p_B)p_A+1\cdot p_Ap_B+P(A)(1-p_B)(1-p_A)$$

solving this equation for $P(A)$ gives

$$P(A)=\frac{p_B}{1-(1-p_A)(1-p_B)}$$

**the probability that both duelists are hit**

using similar logic as in part (A)

$$P(AB)=P(AB|AB^c)P(AB^c)+P(AB|A^cB)P(A^cB)+P(AB|AB)P(AB)+P(AB|A^cB^c)P(A^cB^c)$$

$$=0+0+1\cdot p_Ap_B+P(AB)(1-p_A)(1-p_B) $$

$$P(AB)=\frac{p_Ap_B}{1-(1-p_A)(1-p_B)}$$

**the probability that the duel ends after the nth round of shots**

let $G_i$ be the event that the game goes longer than the $i_th$ duel

the game ends with more than one duel on the condition that

$$P(G_1)=0+0+0+1\cdot (1-p_A)(1-p_B)=(1-p_A)(1-p_B)$$

the game ends with more than two duels on the condition that

$$P(G_2)=(0+0+0+1\cdot (1-p_A)(1-p_B))(0+0+0+1\cdot (1-p_A)(1-p_B))=(1-p_A)^2(1-p_B)^2$$

generalizing this result,

$$P(G_n)=(1-p_A)^n(1-p_B)^n$$

**the conditional probability that the duel ends after the nth round of shots given that A is not hit**

the game ends with more than one duel on the condition that

$$P(G_1|A^c)=P(G_1,AB^c|A^c)P(AB^c)+P(G_1,A^cB|A^c)P(A^cB)+P(G_1,AB|A^c)P(AB)+P(G_1,A^cB^c|A^c)P(A^cB^c)$$

$$=0+0+0+P(G_1,A^cB^c|A^c)(1-p_A)(1-p_B)$$

where

$$P(G_1,A^cB^c|A^c)=\frac{P(G_1,A^cB^c,A^c)}{P(A^c)}$$

$$=\frac{1}{P(A^c)}$$

and $P(A^c)=1-P(A)=1-(p_B(1-p_A)+p_Ap_B)=1-p_B$

therefore

$$P(G_1|A^c)=\frac{(1-p_A)(1-p_B)}{1-p_B}=1-p_A$$

the game ends with more than two duels on the condition that

$$P(G_2|A^c)=(0+0+0+P(G_1,A^cB^c|A^c)(1-p_A)(1-p_B))(0+0+0+P(G_2,A^cB^c|A^c)(1-p_A)(1-p_B))$$

where

$$P(G_1,A^cB^c|A^c)=1$$

and

$$P(G_2,A^cB^c|A^c)=\frac{P(G_2,A^cB^c,A^c)}{P(A^c)}$$

$$=\frac{1}{P(A^c)}$$

and $P(A^c)=1-P(A)=1-(p_B(1-p_A)+p_Ap_B)=1-p_B$

therefore

$$P(G_2|A^c)=(1-p_A)^2(1-p_B)$$

in general then,

$$P(G_n|A^c)=(1-p_A)^n(1-p_B)^{n-1}$$

**the conditional probability that the duel ends after the nth round of shots given that both duelists are hit**

this problem is similar to part (d) so minimal work will be shown,

$$P(G_1,A^cB^c|AB)=\frac{P(G_1,A^cB^c,AB)}{P(AB)}=\frac{1}{P(AB)}$$

where

$$P(AB)=p_Ap_B$$

and hence

$$P(G_1|AB)=\frac{(1-p_A)(1-p_B)}{p_Ap_B}$$

and in general

$$P(G_n|AB)=\frac{(1-p_A)^n(1-p_B)^n}{p_Ap_B}$$

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Asked: 2 years ago