What is the probability of one or both events occurring.
September 18, 2009 12:55 PM Subscribe
What is the probability of one or both events occurring.
Seems like this should be a simple problem. but I have been unable to figure it out. Any help in the form of an answer or a pointer to somewhere that would help me solve it would be very nice.
A procedure is about to occur. There is a 1/100 chance that outcome X will occur, and a 1/200 chance that Y will occur. The outcomes are independent of each other. What is the probability that one or both of the outcomes will occur?
Seems like this should be a simple problem. but I have been unable to figure it out. Any help in the form of an answer or a pointer to somewhere that would help me solve it would be very nice.
A procedure is about to occur. There is a 1/100 chance that outcome X will occur, and a 1/200 chance that Y will occur. The outcomes are independent of each other. What is the probability that one or both of the outcomes will occur?
What is the probability that one or both of the outcomes will occur?
1 minus the probability that neither event will occur. so 1 - (99/100)*(199/200) = .01495 ~ 1.5%
Disclaimer: It's been awhile since I took a probability class.
posted by muddgirl at 12:58 PM on September 18, 2009 [3 favorites]
1 minus the probability that neither event will occur. so 1 - (99/100)*(199/200) = .01495 ~ 1.5%
Disclaimer: It's been awhile since I took a probability class.
posted by muddgirl at 12:58 PM on September 18, 2009 [3 favorites]
Joint probability for independent events is P(X)*P(Y)
So the probability that both X and Y will happen ( P(X and Y) ) is:
1/100 * 1/200 = 1/20000
Probability for at least one of the events to occur is P(X) + P(Y) - P(X)*P(Y)
So the probability that either X or Y will happen ( P(X or Y) ) is:
1/100 + 1/200 - 1/20000 = 0.01495, which is around 1/67
Wikipedia can get you started with this kind of stuff.
posted by demiurge at 1:06 PM on September 18, 2009 [1 favorite]
So the probability that both X and Y will happen ( P(X and Y) ) is:
1/100 * 1/200 = 1/20000
Probability for at least one of the events to occur is P(X) + P(Y) - P(X)*P(Y)
So the probability that either X or Y will happen ( P(X or Y) ) is:
1/100 + 1/200 - 1/20000 = 0.01495, which is around 1/67
Wikipedia can get you started with this kind of stuff.
posted by demiurge at 1:06 PM on September 18, 2009 [1 favorite]
(1/100)*(199/200) + (1/200)*(99/100) + (1/200)*(1/100) = 0.00995 + 0.00495 + 0.00005 = 1.495% chance
Easier way of doing this is to consider that the only way for your condition to NOT be true is if neither event happens. So (99/100) * (199/200) = 0.98505. This is the chance that your condition of one or both events happening will NOT be true. So the rest is the chance that your condition will be true. Thus, 1 - 0.98505 = 0.01495 = 1.495%
posted by fourmajor at 2:45 PM on September 18, 2009
Easier way of doing this is to consider that the only way for your condition to NOT be true is if neither event happens. So (99/100) * (199/200) = 0.98505. This is the chance that your condition of one or both events happening will NOT be true. So the rest is the chance that your condition will be true. Thus, 1 - 0.98505 = 0.01495 = 1.495%
posted by fourmajor at 2:45 PM on September 18, 2009
Response by poster: Great, thank you so much. I will study and, I hope, understand.
posted by charlesminus at 4:42 PM on September 18, 2009
posted by charlesminus at 4:42 PM on September 18, 2009
This thread is closed to new comments.
posted by 256 at 12:58 PM on September 18, 2009