Consider the question. An archer is shooting arrows from various distances. She shoots at a target from 30 yards, 40 yards, and 50 yards. The probability that she hits within the bullseye from 30 yards is 5/6, the probability that she hits within the bullseye from 40 yards is 4/5, and the probability that she hits within the bullseye from 50 yards is 3/4. What is the probability that she hits the bullseye at least once in three shots, one from each distance listed.
Options:
(A) 104/105
(B) 119/120
(C) 139/140
(D) 9/10
(E) 14/15
If the answer is '(B)'; explain the reasoning:
The probability that she hits the bullseye at least once = 1 - (the probability that she misses the bullseye every shot)
The probability that she misses all shots = (prob that she misses from 30 yards)*(prob that she misses from 40 yards)*(prob that she misses from 50 yards)
Prob that she misses from 30 yards = 1 - 5/6 = 1/6; Prob that she misses from 40 yards = 1 - 4/5 = 1/5; Prob that she misses from 50 yards = 1 - 3/4 = 1/4
Thus the probability that she misses all shots = (1/6)*(1/5)*(1/4) = 1/120
Thus the probability that she makes at least one shot = 1 - (1/120) = 119/120