"Seven Points for Black in Snooker".
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 to solve 7 + p = p +7
dependent because your changing
What is the correct answer?
7
7 Points for Black in Snooker
7 Sides on a Fifty Pence Piece...
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
.7
7 Harry Potter Books
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
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.