Peter R. answered 04/10/22
Experienced Instructor in Prealgebra, Algebra I and II, SAT/ACT Math.
Here's another take on the problem. The archer stops when he/she hits the target. Min. 3 tries. How many tries before hitting the target? Prob. of hit = 0.85; prob. of miss = 0.15.
Misses 1st 2 tries, success on 3rd. Prob = 0.15 x 0.15 x 0.85 = 0.0191
Misses 1st 3 tries, success on 4th. Prob = 0.15 x 0.15 x 0.15 x 0.85 = 0.0029
Misses 1st 4 tries, success on 5th (max no. of attempts). Prob = 0.15 x 0.15 x 0.15 x 0.15 x 0.85 = 0.0004
Sum of probability distribution is 0.0224 or 2.24%
I'd appreciate any comments re: this approach.