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If the occurrence of B makes A more likely does the occurrence of A make B more likely?
This is a tricky question, but if you study probability, re-writing it in mathematical notations and using simple definitions and algebra would show you that the answer is yes. We could rewrite the question as: if B occur, A is more likely to occur, Probability of A happening given B happened > probability of A happening P( A given B) > P(A) Is it true that in this case, P(B given A) > P(B)? by definition, P( B given A) = P(A and B) / P( A) = P( A given B) * P(B) / P(A) Since P( A given B) > P(A), then P( A given B)/ P(A) > 1 thus P( B given A) > P(B) and the answer is true, occurence of A makes B more likely. Say A = lung cancer, B = smoking. If someone is smoking, they are more likely to have lung cancer. If someone has lung cancer, they are more likely to be a smoker or have smoked (compared to being a non-smoker).
How do you solve 6 21 b 6 21 b?
how to solve 7 + p = p +7
If P(A)=1/5 and P(B)=3/7 and P(A∩B)=2/17 are A and B independent or dependent?
dependent because your changing
What is P(A and B)(A) 2/7(B) 4/7(C) 5/7(D) 3/7?
What is the correct answer?
What is the ditloid 4 P B?
7
7 p for b in s?
7 Points for Black in Snooker
How is 7 s on a f p p 50 is a b e?
7 Sides on a Fifty Pence Piece...
What is the probability of rolling either a 5 or 6 or 8 or 9 before rolling a 7 with two dice?
Consider the three events: A = rolling 5, 6, 8 or 9. B = rolling 7 C = rolling any other number. Let P be the probability of these events in one roll of a pair of dice. Then P(A) = P(5) + P(6) + P(8) + P(9) = 18/36 = 1/2 P(B) = P(7) = 6/36 = 1/6 and P(C) = 1 - [P(A) + P(B)] = 1/3 Now P(A before B) = P(A or C followed by A before B) = P(A) + P(C)*P(A before B) = 1/2 + 1/3*P(A before B) That is, P(A before B) = 1/2 + 1/3*P(A before B) or 2/3*P(A before B) = 1/2 so that P(A before B) = 1/2*3/2 = 3/4
Suppose A and B are independent events. If p a 0.3 and what is?
.7
What does this diltoid mean 7 H P B?
7 Harry Potter Books
If A and B are independent events then are A and B' independent?
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
What is the product rule and the sum rule of probability?
Sum Rule: P(A) = \sum_{B} P(A,B) Product Rule: P(A , B) = P(A) P(B|A) or P(A, B)=P(B) P(A|B) [P(A|B) means probability of A given that B has occurred] P(A, B) = P(A) P(B) , if A and B are independent events.
