Black birds baked in a pie.
80 divided by four equals 20.
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
The answer is negative four BECAUSE... 20/5 is POSITIVE four 20/-5 is a NEGATIVE four because a positive divided by a negative is a negative. Easy way to remember negs/pos: n * p = n p* n = n p * p = p n * n = p n / p = n p / n = n p / p = p n / n = p There are always two ways to get a positive, and two ways to get a negative. Very simple.
If at least three flips are tails, there are two scenarios where we can obtain exactly four tails in five flips. Either the first four flips are tails and the last flip is heads, or the first flip is heads and the next four flips are tails. Each scenario has a probability of 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32. Therefore, the probability of obtaining exactly four tails in five flips if at least three are tails is 1/32 + 1/32 = 1/16.
four and twenty blackbirds baked in a pie - from the nursery rhyme 'sing a song of sixpence'
My best guess is the in should be and, so it would be Four and Twenty Black Birds in a Pie
Black birds baked in a pie.
You tell me :P
Area = square root of {s1(s1-a)(s1-b)(s1-p)} + square root of {s2(s2-c)(s2-d)(s2-p)} where a,b,c and d are the four sides of the quadrilateral, p is the diagonal separating the sides a,b from c,d, and s1 = (a+b+p)/2 and s2 = (c+d+p)/2
80 divided by four equals 20.
The question asks for the probability of an even card OR a red card. The term "OR" is key since this is not the same as the probability of drawing an even card and a red card, that is to say an even red card. GIven any two events, A and B P(A or B)=P(A)+P(B)-P(A and B) IF A and B are mutually exclusive, then P(A and B)=0 and this equation becomes P(A)+P(B) However, they are NOT in this case. So let A be the probability the card is even and B the probability it is red. P(A)=20/52 since J, K and Q are neither even nor odd (20=(52-12)/2)) P(B)=26/52 since half the cards are red. P(A and B) is the probability that a card is red AND even. We have 20 even cards, half of them are red and half are black so the odds are 10/52 of being red and even. P (A or B)=20/52+26/52-10/52=9/13
P = peaches, B = plumbs P + B = 32 P + 2B = 52 Subtract the first line from the second to get B = 20 So he bought 20 plumbs.
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
Four and Twenty Blackbirds Baked in a Pie.
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
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.