John is playing a game of darts. The probability that he throws a dart into the center of the dart board (the Bull’s eye) is 1/10. The probability that he throws the dart into the 10-point ring is 3/10.
What is the probability that he either hits a Bull’s eye or scores 10 points?

Question

John is playing a game of darts. The probability that he throws a dart into the center of the dart board (the Bull’s eye) is 1/10. The probability that he throws the dart into the 10-point ring is 3/10.
What is the probability that he either hits a Bull’s eye or scores 10 points?

xenesis98 · Answer

a. 1/3
b. 2/3
c.  3/5
d. 2/5
e. 1/4

directrix · Answer

Probability that he either hits a Bull’s eye or scores =

P(Bull's Eye) + P(10 Point Ring) = 1/10 + 3/10 = 

Because the dart cannot hit the Bull's Eye and the 10 Point Ring at the same time, the events are mutually exclusive.

You can just add the individual probabilities to get the probability that he hits a Bull’s eye or scores 10 points.

1/10 + 3/10 =     @xenesis98

xenesis98 · Answer

2/5

directrix · Answer

4/10 = 2/5   Correct.