# Probability Hitting a Target

Problem:

• Jeff hits a target independently of Bob with probability $$0.7$$.

• Bob hits a target independently of Jeff with probability $$0.4$$.

Say they both shoot at a target at the same time.

What is the probability that the target is hit?

What is the probability that Jeff hit the target given that the target was hit?

My attempt: For the target to be hit we have three possibilities. Either they both hit the target, only jeff hits the target or only bob hits the target so we get

$$(.4)(.7)+(0.3)(0.4)+(0.7)(0.6) =.82$$

To find the probability that Jeff hit the target given that the target was hit, would it be $$\frac{(.4)(.7)+(0.7)(0.6)}{.82}=\frac{.7}{.82}=\frac{70}{82} .$$

Does this work? Thanks

• Leaving an answer as $\frac {.4}{.82}$ is bad form. But otherwise the calculation looks good.
– lulu
Oct 17, 2019 at 17:39
• thanks, we werent allowed calculators. Oct 17, 2019 at 17:39
• Sure, but even so. $\frac {.4}{.82}=\frac {40}{82}=\frac {20}{41}$ can be done without calculators.
– lulu
Oct 17, 2019 at 17:41
• Oh, sorry. Your calculation is for Bob, not Jeff. The $.3\times .4$ term is incorrect.
– lulu
Oct 17, 2019 at 17:42
• Yes but you want the probability that Jeff hits it, not Bob. Oct 17, 2019 at 17:49

I believe for your second answer should be $$\frac{(.4)(.7)+(0.7)(0.6)}{0.82}$$
$$P(J|T)=\frac{P(T|J)P(J)}{P(T)}$$
$$P(T|J)$$ is 1. $$P(J)$$ is $$(.4)(.7)+(0.7)(0.6)$$ and you calculated $$P(T)$$ correctly in the first part
• @elcharlosmaster you said Jeff is the one who hits the target with probability $0.7$, right? Bob is the one who hits with probability $0.4$. Oct 17, 2019 at 17:44