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?
pr each misses; (.4)(.7)=.28
pr both hits=(.6)(.3)=.18
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
To find the probability that the game is a tie, we need to determine the probability that both Mike and Dave either hit the target or miss the target. In other words, we need to find the combined probability of both hitting the target and both missing the target.
The probability that Mike hits the target is 0.6, and the probability that Dave hits the target is 0.3. To find the combined probability of both hitting the target, we multiply these probabilities: 0.6 * 0.3 = 0.18.
Similarly, the probability that Mike misses the target is 1 - 0.6 = 0.4, and the probability that Dave misses the target is 1 - 0.3 = 0.7. To find the combined probability of both missing the target, we multiply these probabilities: 0.4 * 0.7 = 0.28.
Now, to find the probability of a tie, we need to find the combined probability of both hitting the target and both missing the target and add them together: 0.18 + 0.28 = 0.46.
Therefore, the probability that the game is a tie is 0.46, or 46%.