Calculating conditional probabilities - random permutations. about the letters (a, b, c, d, e, f, g) are put in a random order. each permutation is equally likely. define the following events: a: the letter b falls in the middle (with three before it and three after it) b: the letter c appears to the right of b, although c is not necessarily immediately to the right of b. for example, "agbdcef" would be an outcome in this event. c: the letters "def occur together in that order (e. g. "gdefbca") calculate the probability of each individual event. that is, calculate p(a), p(b), and p(c). what is p(aic)? (c) what is p(bic)? what is p(aib)? (e) which pairs of events among a, b, and c are independent? feedback?

A="b is in the middle"

B="c is to the right of b"

C="The letter def occur together in that order"

a) b can be in 7 places, but only one is the middle. So, P(A)=1/7

b) X=i, "b is in the i-th position"

Y=j, "c is in the j-th position"

P(B)=1/2

c) X=i, "d is in the i-th position"

Y=j, "e is in the j-th position"

Z=k, "f is in the i-th position"

P(C)=1/42

P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420

P(B∩A)=3*(1/7*1/6)=1/14

P(A|C)=P(A∩C)/P(C)=(1/420)/(1/42)=1/10

P(B|C)=P(B∩C)/P(C)=(1/420)/(1/42)=1/10

P(A|B)=P(B∩A)/P(B)=(1/14)/(1/2)=1/7

P(A∩B)=1/14

P(A)P(B)=(1/7)*(1/2)=1/14

A and B are independent

P(A∩C)=1/420

P(A)P(C)=(1/7)*(1/42)=1/294

A and C aren't independent

P(B∩C)=1/420

P(B)P(C)=(1/2)*(1/42)=1/84

B and C aren't independent

l.ln.lln.k.nkl

step-by-step explanation:

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

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