Calculating conditional probabilities - random permutations. About The letters (a, b, c, d, e, f, g) are put in a random order.

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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?

Mathematics

1 answer:

AnnZ [12.3K] · Answer 1 · 2026-03-19T16:55:10+03:00

A="b is situated in the center"

B="c lies to the right of b"

C="The letters def occur sequentially in that arrangement"

a) b can occupy 7 positions; however, only one of these is the center. Therefore, P(A)=1/7

b) Let X=i; "b holds the i-th position"

Y=j; "c occupies the j-th position"

$P(B)=\displaystyle\sum_{i=1}^{6}(P(X=i)\displaystyle\sum_{j=i+1}^{7}P(Y=j))=\displaystyle\sum_{i=1}^{6}\frac{1}{7}(\displaystyle\sum_{j=i+1}^{7}\frac{1}{6})=\frac{1}{42}\displaystyle\sum_{i=1}^{6}(\displaystyle\sum_{j=i+1}^{7}1)=\frac{6+5+4+3+2+1}{42}=\frac{1}{2}$

P(B)=1/2

c) Let X=i; "d holds the i-th position"

Y=j; "e occupies the j-th position"

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

$P(C)=\displaystyle\sum_{i=1}^{5}( P(X=i)P(Y=i+1)P(Z=i+2))=\displaystyle\sum_{i=1}^{5}(\frac{1}{7}\times\frac{1}{6}\times\frac{1}{5})=\frac{1}{210}\displaystyle\sum_{i=1}^{5}(1)=\frac{1}{42}$

P(C)=1/42

P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420

$P(B\cap C)=\displaystyle\sum_{i=1}^{3} P(X=i)P(Y=i+1)P(Z=i+2)\displaystyle\sum_{j=i+3}^{6}P(V=j)P(W=j+1)=\displaystyle\sum_{i=1}^{3}\frac{1}{6}\frac{1}{7}\frac{1}{5}(\displaystyle\sum_{j=1+3}^{6}\frac{1}{4}\frac{1}{3})=1/420$