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4 and 20 Blackbirds baked in a pie schwazoo

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Q: What is four and 20 b b in a p?
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4 and 20 b b b in a P?

Black birds baked in a pie.


What is 80 divided by b equals 20 for b equals 4?

80 divided by four equals 20.


How do you find p(b) when p(a) is 23 p(ba) is 12 and p(a U b) is 45 and is a dependent event?

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


What is twenty divided by negative five?

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.


What is the probability of obtaining exactly four tails in five flips of a coin if at least three are tails?

We need to determine the separate event. Let A = obtaining four tails in five flips of coin Let B = obtaining at least three tails in five flips of coin Apply Binomial Theorem for this problem, and we have: P(A | B) = P(A ∩ B) / P(B) P(A | B) means the probability of "given event B, or if event B occurs, then event A occurs." P(A ∩ B) means the probability in which both event B and event A occur at a same time. P(B) means the probability of event B occurs. Work out each term... P(B) = (5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0 It's obvious that P(A ∩ B) = (5 choose 4)(½)4(½) since A ∩ B represents events A and B occurring at the same time, so there must be four tails occurring in five flips of coin. Hence, you should get: P(A | B) = P(A ∩ B) / P(B) = ((5 choose 4)(½)4(½))/((5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0)

Related questions

What does 4 and 20 b b b in a p mean?

four and twenty blackbirds baked in a pie - from the nursery rhyme 'sing a song of sixpence'


What does 4 in 20 B B in a P stand for?

My best guess is the in should be and, so it would be Four and Twenty Black Birds in a Pie


4 and 20 b b b in a P?

Black birds baked in a pie.


Has four letters starting with I AND ENDING with b?

You tell me :P


How do you find an area of a four sided diagram?

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


What is 80 divided by b equals 20 for b equals 4?

80 divided by four equals 20.


If you draw a card from a normal deck of cards what is the probability that you will draw an even number or red card?

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


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


32 plumbs and peaches for 52 dollars plums are2dollar and peaches are 1 dollar how many plumbs did he buy?

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.


Find a phrase containing 4 and T B B in a P?

Four and Twenty Blackbirds Baked in a Pie.


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.


How do you find p(b) when p(a) is 23 p(ba) is 12 and p(a U b) is 45 and is a dependent event?

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