The probability that it will be rainy on Monday is 7/10. The probability that it will be windy on Monday is 6/10. These events are independent. What is the probability that it will be rainy or windy on Monday?
2007-07-24 07:03:57 · 8 answers · asked by t.wes87 2 in Science & Mathematics ➔ Mathematics
Since these are independent events, you need to consider this situation:
Prob(Wind or Rain) = Prob(Rain) + Prob(Wind) - Prob(Rain and Wind)
Prob(Wind or Rain) = 0.7 + 0.6 - 0.7*0.6 = 1.3 - .42 = 0.88
So there is 88% probability that it will be rainy or windy on Monday.
The key to this question (and to not get non-sensical >100% answers) is to realize that since it can be both rainy and windy, you don't want to count the probability twice (which is why you subtract the cases where there's both rain and wind)
2007-07-24 07:09:24 · answer #1 · answered by sharky.mark 4 · 1⤊ 3⤋
Answer = 0.7 + (1-0.7)*0.6 = 0.6 + (1-0.6)*0.7 = 0.88
7 of 10 Mondays are rainy. Of the remaining 3 Mondays, how many will be windy? Since the events are independent, we can apply the same probability (6/10) to the remaining 3 Mondays which makes it 3*6/10=1.8 Mondays. Therefore, a total of 8.8 Mondays out of 10 will be rainy or windy giving a probability of 0.88.
We would expect that we should be able to start with the number of windy Mondays and sum the number of rainy Mondays out of the non-windy Mondays and get the same answer. Yes, it is true. It follows from the fact that x+(1-x)*y = y+(1-y)*x.
2007-07-24 14:49:03 · answer #2 · answered by gandev 1 · 0⤊ 0⤋
The probability that it will not be windy = 3/10
The probability that it will not be rainy = 4/10
So, the probability that it will not be rainy and windy = 12/100
So, the probability that it will be rainy or windy = 88/100
Note: You cannot directly add or multiply the positive probabilities in this case.
2007-07-24 14:11:40 · answer #3 · answered by Swamy 7 · 0⤊ 0⤋
P(rainy or windy)=1-P(not rainy and not windy)
=1-P(not rainy)*P(not windy)
=1-(3/10)*(4/10)
=88/100
=88%
2007-07-24 14:11:29 · answer #4 · answered by Anonymous · 2⤊ 0⤋
p( of no rain and wind) = 3/10 * 4/10 = 12/100
therefore p(of at least wind or rain) = 1 - 12/100 = 88/100
2007-07-24 14:10:22 · answer #5 · answered by Anonymous · 3⤊ 0⤋
you must live in Seattle because there is a 130% chance of it either being rainy or windy
2007-07-24 14:09:50 · answer #6 · answered by walsh_patr 3 · 1⤊ 3⤋
I believe its 7/10 + 6/10, which equals 13/10
2007-07-24 14:07:53 · answer #7 · answered by Ben H 1 · 0⤊ 6⤋
P(rain) = 0.7
P(wind) = 0.6
P(rain OR wind) = P(rain) + P(wind) - P(rain AND wind)
=0.7 + 0.6 - 0.6*0.7
= 0.88
2007-07-27 15:06:05 · answer #8 · answered by Merlyn 7 · 0⤊ 0⤋