Mike and Dave play a game in which each independently throws a dart at a target. Mike hits the target with probability 0.6, while Dave hits the target with probability 0.3. Mike wins the game if he hits the target and Dave misses. Dave wins the game if he hits the target and Mike misses. Otherwise the game is a tie. What is the probability that the game is a tie?
3 answers
pr each misses; (.4)(.7)=.28
pr both hits=(.6)(.3)=.18
Pr tie=prob both hits+pr each misses=.28+.18=.46
Pr tie=prob both hits+pr each misses=.28+.18=.46
There are four possible results
A. Mike and Dave both miss
B. Mike hit, Dave miss
C. Dave hit, Mike miss
D. both hit
p mike hits = .6 so misses = .4
p dave hits = .3 so misses = .7
So
p(A) = .4 * .7
p(B) = .6 * .7
p(C) = .3 * .4
p(D) = .6 * .3
We want p(A) + p(D)
= .28 + .18 = .46
A. Mike and Dave both miss
B. Mike hit, Dave miss
C. Dave hit, Mike miss
D. both hit
p mike hits = .6 so misses = .4
p dave hits = .3 so misses = .7
So
p(A) = .4 * .7
p(B) = .6 * .7
p(C) = .3 * .4
p(D) = .6 * .3
We want p(A) + p(D)
= .28 + .18 = .46
1 answer