Brandon is an amateur marksman. When he takes aim at a particular target on the shooting range, there is a 0.10.10, point, 1 probability that he will hit it. One day, Brandon decides to attempt to hit 101010 such targets in a row.
Assuming that Brandon is equally likely to hit each of the 101010 targets, what is the probability that he will hit at least one of them?
+36916 Corrected....sometimes the numbers 'triple'
Brandon is an amateur marksman. When he takes aim at a particular target on the shooting range, there is a 0.10. probability that he will hit it. One day, Brandon decides to attempt to hit 10 such targets in a row.
Assuming that Brandon is equally likely to hit each of the 101010 targets, what is the probability that he will hit at least one of them?
+363 Ahhh... ok
so a 10percent chance to hit one target.
so if he tries ten times, there is a 10x10=100 percent chance that he will hit a target
so 100 percent.
Hmmmmm maybe this:
He has a .9 chance of MISSING ...in 10 tries
.910 = .348
chance of hitting then becomes 1 - .348 = .652 65 % chance of hitting at LEAST one
+363 WAITWAITWAIT
In this situation it is much easier to calculate the probability of the event we are looking for (he hits at least one target) by calculating the probability of its complement (he misses every target), and subtracting from 1.
In other words, we can use this strategy:
P(at least one hit)=1-P(miss all 10)
Calculations:
P(at least one hit)
=1-P(miss all 10)
=1-(0.9)^10
≈1-0.349
≈0.65
Answer:
P(at least one hit)≈0.65
oops....
