# Three riflemen take one shot each at the same target. The probability of the first riflemen hitting the target is 0.4, the probability of the second rifleman hitting the target is 0.5 and the probability of the third rifleman hitting the target is 0.8. The probability that exactly two of them hit the target is _______.

So,

Probability of the first riflemen missing the target=$1-0.4=0.6$,

Probability of the second riflemen missing the target=$1-0.5=0.5$,

Probability of the third riflemen missing the target=$1-0.8=0.2$.

Now,

Probability that exactly two of them hit the target = P(first rifle man and second rifleman hit the target and third rifleman misses)+ P(first rifle man and third rifleman hit the target and second rifleman misses)+ P(third rifle man and second rifleman hit the target and first rifleman misses)

Therefore,

Probability that exactly two of them hit the target = $\left(0.4\right)\left(0.5\right)\left(0.2\right)+\left(0.4\right)\left(0.5\right)\left(0.8\right)+\left(0.6\right)\left(0.5\right)\left(0.8\right)=\frac{40}{1000}+\frac{160}{1000}+\frac{240}{1000}=\frac{440}{1000}=0.44$

Hence, the probability that exactly two of them hit the target is _0.44______.

**
**