Question

The probability that A hits a target is 1 / 4 and the probability that B hits a target is 1/3 .
 If each fires once and the target is hit only once, what is the probability that A hits the target?

Comments

Is the answer 0.125 ?

---

is it 1/4 * 2/3 ??

---

ans given is 1/3

Answer 1

$Prob(\text{A hits target = A}) = \frac{1}{4}$

$Prob(\text{B hits target = B}) = \frac{1}{3}$

Given ,that Target is hit only once:

$ Prob(\text{Target is hit only once = O}) = \frac{1}{4}*\frac{2}{3}+\frac{3}{4}*\frac{1}{3} = \frac{10}{24} $

Probability, that A hits the target, given that the target was hit only once:

$Prob(A|O) = \frac{Prob(A \cap O)}{Prob(O)} = \frac{1}{6}*\frac{24}{10} = \frac{4}{10} = 0.4 $

Comments

i also think the same but ans given in Seymore Lipschutz is 1/3

---

I see. I will ask someone who knows this area. Let see what he says. But i feel this is correct and there is a chance that the answer is incorrect.

---

4/10 or 2/5 is correct.

---

yes ans given in book must be incorrect  

              2/5 is right

Answer 2

p(A hits)=1/4,P(A do not hit)=3/4;P(B hits target)=1/3;P(B do not hit)=2/3

 

P( A hits the target) =P(B do not hit)*P(A hits)=2/3*1/4=1/6