Eats Dirt said:Homework Statement
If event A equals event B, then the probability of their intersection is 1. True or False?
Apparently the correct answer is False.
The Attempt at a Solution
If A=B then they should overlap entirely and their intersection should be 1? The only way I see this working is if A is a subset of B and therefore they do not overlap completely, but when the question states "equals" I would think this means they are the same.
Yes, I agree that A = B in this sense because there are two options, heads or tails - getting a head and not getting a tail is the same thing. By definition the intersection would be those elements that are in common between the events, in this case getting heads. It is 100% certain in terms of probability because the elements of each event (getting heads or not getting tails) are the same and thus their P(intersection) = 1. So isn't the statement true?Ray Vickson said:Suppose that in one toss of a coin we have A ={get heads} and B = {do not get tails}. Do you agree that A=B? What is their intersection? Why do you think that their intersection is 100% certain?
Eats Dirt said:Yes, I agree that A = B in this sense because there are two options, heads or tails - getting a head and not getting a tail is the same thing. By definition the intersection would be those elements that are in common between the events, in this case getting heads. It is 100% certain in terms of probability because the elements of each event (getting heads or not getting tails) are the same and thus their P(intersection) = 1. So isn't the statement true?
But it is asking about the probability of their "Intersection" so shouldn't the intersection between the two overlap completely and be 1?Ray Vickson said:So, I can be 100% sure to get a "head" in a coin-toss just by describing it in two ways? If I say "get heads", that has probability 1/2, and if I say "do not get tails" that has probability 1/2 also, but if I say it in two different ways it suddenly has probability 1?
Eats Dirt said:But it is asking about the probability of their "Intersection" so shouldn't the intersection between the two overlap completely and be 1?
Their intersection would be a subset of S say, ##Sub = \{ H \}.##Ray Vickson said:That is why I asked you to tell me what is the intersection of the events A={get heads} and B = {do not get tails}. The intersection ##A \cap B## is some subset of the sample space ##S = \{ H,T \}.## What IS that subset? Do not tell me in words; actually display the subset.
ok i got it thank youEats Dirt said:Their intersection would be a subset of S say, ##Sub = \{ H \}.##
Eats Dirt said:ok i got it thank you
The probability of intersection refers to the likelihood of two or more events occurring simultaneously or together. In other words, it is the chance that both events A and B will happen at the same time.
The probability of intersection is calculated by multiplying the individual probabilities of each event. This can be represented as P(A and B) = P(A) x P(B). For example, if the probability of event A is 0.6 and the probability of event B is 0.5, then the probability of A and B occurring together is 0.6 x 0.5 = 0.3.
No, the probability of intersection cannot be greater than 1. This is because the maximum probability of an event occurring is 1, and when two or more events are combined, the probability cannot exceed this limit.
If two events are independent, meaning that the occurrence of one event does not affect the probability of the other event, then the probability of intersection is simply the product of the individual probabilities. However, if the events are dependent, meaning that the occurrence of one event affects the probability of the other, then the calculation is more complex and involves conditional probabilities.
The probability of intersection is an important concept in statistics as it allows us to understand the likelihood of multiple events occurring together. It is also used in various statistical calculations, such as the calculation of joint probabilities and conditional probabilities.