Question

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?

Answer 1

Answer:

P(A)=1/7

P(B)=1/2

P(C)=1/42

P(A|C)=1/10

P(B|C)=1/10

P(A|B)=1/7

A and B are independent

A and C aren't independent

B and C aren't independent

Step-by-step explanation:

A="b falls in the middle"

- b can fall in seven possible places, but only one is the middle. So, P(A)=1/7

B="c falls to the right of b"

X=i means "b falls in the i-th position"

Y=j means "c falls in the j-th position"

if b falls in the first place, c can fall in the 2nd, 3rd, 4th, 5th, 6th or 7th place.

if b falls in the 2nd place, c can fall in the 3rd, 4th, 5th, 6th or 7th place

 ...

If b falls in the 6th place, c can fall in the 7th place

then:

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

- if d falls in the 1st place, e falls in the 2nd and f in the 3rd place

- if d falls in the 2nd place, e falls in the 3rd and f in the 4th place

- if d falls in the 3rd place, e falls in the 4th and f in the 5th place

- if d falls in the 4th place, e falls in the 5th and f in the 6th place

- if d falls in the 5th place, e falls in the 6th and f in the 7th place

X=i means "d falls in the i-th position"

Y=j means "e falls in the j-th position"

Z=k means "f falls in the k-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(A|C)=P(A∩C)/P(C)=?

A∩C:

- d falls in the 1st place, e in the 2nd, f in the 3rd and b in the 4th place

- b falls in the 4th place, d in the 5th place, e in the 6th, f in the 7th place

P(A∩C)=2*(1/7*1/6*1/5*1/4)=1/420

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

P(B|C)=P(B∩C)/P(C)=?

X=i means "d falls in the i-th position"

Y=j means "e falls in the j-th position"

Z=k means "f falls in the k-th position"

V=k means "b falls in the k-th position"

W=k means "c falls in the k-th position"

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

$P(B\cap C)=\displaystyle\sum_{i=1}^{3} \frac{1}{7}\times\frac{1}{6}\times\frac{1}{5}(\displaystyle\sum_{j=i+3}^{6}\frac{1}{4}\times\frac{1}{3})=\frac{1}{2520}\displaystyle\sum_{i=1}^{3} \displaystyle\sum_{j=i+3}^{6}1=\frac{1}{420}$

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

P(A|B)=P(B∩A)/P(B)=?

B∩A:

- b falls in the 4th place and c in the 5th

- b falls in the 4th place and c in the 6th

- b falls in the 4th place and c in the 7th

P(B∩A)=3*(1/7*1/6)=1/14

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

If one event is independent of another, P(X∩Y)=P(X)P(Y)

So:

P(A∩B)=1/14=(1/7)*(1/2)=P(A)P(B), A and B are independent

P(A∩C)=1/420≠(1/7)*(1/42)=1/294=P(A)P(C), A and C aren't independent

P(B∩C)=1/420≠(1/2)*(1/42)=1/84=P(A)P(C), B and C aren't independent