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There are symbols missing from your question which I cam struggling to guess and re-insert. p(a) = 2/3 p(b ??? a) = 1/2 p(a ∪ b) = 4/5 p(b) = ? Why use the set notation of Union on the third given probability whereas the second probability has something missing but the "sets" are in the other order, and the order wouldn't matter in sets. There are two possibilities: 1) The second probability is: p(b ∩ a) = p(a ∩ b) = 1/2 → p(a) + p(b) = p(a ∪ b) + p(a ∩ b) → p(b) = p(a ∪ b) + p(a ∩ b) - p(a) = 4/5 + 1/2 - 2/3 = 24/30 + 15/30 - 20/30 = 19/30 2) The second and third probabilities are probabilities of "given that", ie: p(b|a) = 1/2 p(a|b) = 4/5 → Use Bayes theorem: p(b)p(a|b) = p(a)p(b|a) → p(b) = (p(a)p(b|a))/p(a|b) = (2/3 × 1/2) / (4/5) = 2/3 × 1/2 × 5/4 = 5/12
Let A= drawing a ten B= drawing a heart P(A or B) = P(A) + P(B) - P(A & B) A & B denotes a ten of hearts. P( ten or a heart)= 4/52 + 13/52 -1/52 =16/52 = 4/13
Probability (P) of A or B is: P(A) + P(B) - P(A and B). Apply to the question is: P(Q) + P(Club) - P (Q&Club) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13 or .3077 or 30.77%.
7
The probability of A is denoted P(A) and the probability of B is denoted P(B). P(A or B) = P(A) + P(B) - P(A and B). Say P(A) = Probability of drawing a heart, which is 13/52. Say P(B) = Probability of drawing a three, which is 4/52. We now have to determine P(A and B) which is the probability of a heart and a three, which is 1/52. We now can determine the probability of drawing a heart or a three which is 13/52 + 4/52 - 1/52 = 16/52 = 4/13.
4 l on a c = 4 legs on a chair, 4 f on b b is 4 faces on big ben------- so what is 10 p in g
Black birds baked in a pie.
P(A)+P(B)+P(C) = 1 so P(A)+P(A)+2*P(A) = 1 => P(A) = 1/4 and therefore, P(A') = 1 - P(A) = 3/4
4 and twenty blackbirds baked in a pie.
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
4
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.