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p(A and B) = p(A) x p(B) for 2 independent events p(A and B and ...N) = p(A) x p(B) x p(C) x ...x p(N) In words, if these are all independent events, find the individual probabilities if each and multiply them all together.
If two events are disjoint, they cannot occur at the same time. For example, if you flip a coin, you cannot get heads AND tails. Since A and B are disjoint, P(A and B) = 0 If A and B were independent, then P(A and B) = 0.4*0.5=0.2. For example, the chances you throw a dice and it lands on 1 AND the chances you flip a coin and it land on heads. These events are independent...the outcome of one event does not affect the outcome of the other.
If the probability of A is p1 and probability of B is p2 where A and B are independent events or outcomes, then the probability of both A and B occurring is p1 x p2. See related link for examples.
Two events are said to be independent if the result of the second event is not affected by the result of the first event. Some common ways to teach this are to perform simulations with coin flips.Students need to understand that if A and B are independent events, the probability of both events occurring is the product of the probabilities of the individual events.Students can predict and then observe probabilities of a fixed number of heads or tails.This lets then see the ideas in action.
No, independence means they are not related. Mutually exclusive means they cannot occur at the same time.
Yes.
Given two events, A and B, Pr(A and B) = Pr(A)*Pr(B) if A and B are independent and Pr(A and B) = Pr(A | B)*Pr(B) if they are not.
p(A and B) = p(A) x p(B) for 2 independent events p(A and B and ...N) = p(A) x p(B) x p(C) x ...x p(N) In words, if these are all independent events, find the individual probabilities if each and multiply them all together.
if P(A)>0 then P(B'|A)=1-P(B|A) so P(A intersect B')=P(A)P(B'|A)=P(A)[1-P(B|A)] =P(A)[1-P(B)] =P(A)P(B') the definition of independent events is if P(A intersect B')=P(A)P(B') that is the proof
when the occurance of an event B is not affected by the occurance of event A than we can say that these events are not dependent with each other
the circles do not overlap at all.
first prove *: if A intersect B is independent, then A intersect B' is independent. (this is on wiki answers) P(A' intersect B') = P(B')P(A'|B') by definition = P(B')[1-P(A|B')] since 1 = P(A) + P(A') = P(B')[1 - P(A)] from the first proof * = P(B')P(A') since 1 = P(A) + P(A') conclude with P(A' intersect B') = P(B')P(A') and is therefore independent by definition. ***note*** i am a student in my first semester of probability so this may be incorrect, but i used the first proof* so i figured i would proof this one to kinda "give back".
P(A given B)*P(B)=P(A and B), where event A is dependent on event B. Finding the probability of an independent event really depends on the situation (dart throwing, coin flipping, even Schrodinger's cat...).
If two events are disjoint, they cannot occur at the same time. For example, if you flip a coin, you cannot get heads AND tails. Since A and B are disjoint, P(A and B) = 0 If A and B were independent, then P(A and B) = 0.4*0.5=0.2. For example, the chances you throw a dice and it lands on 1 AND the chances you flip a coin and it land on heads. These events are independent...the outcome of one event does not affect the outcome of the other.
apex XD 0.140.14
If the probability of A is p1 and probability of B is p2 where A and B are independent events or outcomes, then the probability of both A and B occurring is p1 x p2. See related link for examples.
No. A can be independent of both B and C and this doesn't give use any information about the relationship between B and C.