Sanity check for my logic in a "tree" like problem There is a bomber who is sent to destroy a target. Prob(bomber reach target)=$0.6$ Prob(BOMB hit target)=$0.5$ Prob(target destroyed by bomb)=$0.8$ We send 1 bomber which carry 5 bombs to a target, what is the chance that it destroys it? What I have done is as follows: Each bomb has $0.5*0.8=0.4$ chance to hit & destroy a target which means $0.6$ chance to leave the target unaffected. So $5$ bombs has $(0.6)^5=\dfrac{243}{3125}$ chance to not affect the target. So the prob that 1 bomber with 5 bombs won't kill a target is: $0.4 + 0.6*\dfrac{243}{3125}=0.4466$ This means that the chance that it will destroy the target is $1-0.4466=0.5533$ Is that the right way of doing it? Thanks in advance

$\color{green}\checkmark$ That was a little long winded, but yes, the correct approach.

You're after the probability that the bomber reaches the target and that the five bombs don't all fail to hit and destroy it.

$\mathsf P(\textsf{reach})~\Big(1-\big(1-\mathsf P(\textsf{hit}\mid\textsf{reach})~\mathsf P(\textsf{destroy}\mid\textsf{hit})\big)^5\Big)\\\= (0.6)\Big(1-\big(1-(0.5)(0.8)\big)^5\big)\\\=0.553344$