One of the three shooters is called to the line of fire and makes two shots. The probability of hitting the target for the first shooter – 0.3, for the second – 0.5, and for the third – 0.8. The target is not hit. Find the probability that the shots were made by the first shooter.
The probability for the first shooter to be chosen is 1/3. The probability to miss 1 time is 0.7.
So the answer should be (1/3) * 0.7 * 0.7.
Do I make a mistake?
Best Answer
If the first shooter is selected, the probability of not hitting the target at all is $(1-0.3)^2=0.49$. The same probabilities for the second and third are $(1-0.5)^2=0.25$ and $(1-0.8)^2=0.04$. Then use the law of total probability to find the final answer as $$\frac{0.49}{0.49+0.25+0.04}=\frac{49}{78}$$