Arranging the letters in Mississippi. Have the answer, can someone explain it?

The eleven letters of the word MISSISSIPPI are scrambled and then arranged in some order.

(1) What is the probability that the four I's are consecutive letters in the resulting arrangement?

(2) What is the conditional probability that the four I's are consecutive (event A), given B , where B is the event that arrangement starts with M and ends with S?

(3) What is conditional probability of A, as deŻned above, given C, where C is the event that the arrangements ends with four consecutive S's?

I have the answers and have been sitting here trying to figure out the reasoning behind it, but I have had no luck. Can anybody help me out?

The answers are as follows

1.)

8*[(7!)/(1!*4!*2!)]/[(11!)/1!*4!*4!*2!)] = 4/165

2.)

6*[C(5,2)]/[(9!)/(4!*3!*2!)] = 1/21

3.)

4*[C(3,2)]/[(7!)/(1!*2!*4!)] = 4/35

Arrangements with like items

Hello babbagandu Quote:

Originally Posted by

**babbagandu** The eleven letters of the word MISSISSIPPI are scrambled and then arranged in some order.

(1) What is the probability that the four I's are consecutive letters in the resulting arrangement?

(2) What is the conditional probability that the four I's are consecutive (event A), given B , where B is the event that arrangement starts with M and ends with S?

(3) What is conditional probability of A, as deŻned above, given C, where C is the event that the arrangements ends with four consecutive S's?

I have the answers and have been sitting here trying to figure out the reasoning behind it, but I have had no luck. Can anybody help me out?

The answers are as follows

1.)

8*[(7!)/(1!*4!*2!)]/[(11!)/1!*4!*4!*2!)] = 4/165

2.)

6*[C(5,2)]/[(9!)/(4!*3!*2!)] = 1/21

3.)

4*[C(3,2)]/[(7!)/(1!*2!*4!)] = 4/35

Generalising Plato's result gives the number of arrangements of $\displaystyle n$ items, where there are $\displaystyle a_1$ alike of the first kind, $\displaystyle a_2$ alike of the second kind, ... and so on, which is: $\displaystyle \frac{n!}{a_1!a_2!...}$

1) To get the probability where all four I's are together, treat the block of 4 I's as a single item. So we have 8 items altogether, with 4 alike of the first kind (S) and 2 alike of the second kind (P). Using the general result above, then, the number of arrangements is$\displaystyle \frac{8!}{4!2!}$

So the probability of this is$\displaystyle \frac{8!}{4!2!}\div\frac{11!}{4!4!2!}=\frac{3}{165 }$

2) If the arrangement starts with M and ends with S, the remaining 9 letters have 4 alike of the first kind (I), 3 alike of the second kind (S) and 2 alike of the third kind (P). Using the general result above, then, there are therefore $\displaystyle \frac{9!}{4!3!2!}$ arrangements of these letters.

If we treat the 4 I's as a single block (as in question (1)), we have $\displaystyle \frac{6!}{3!2!}$ arrangements. So the probability is$\displaystyle \frac{6!}{3!2!}\div\frac{9!}{4!3!2!}=\frac{1}{21}$

You may wonder why the answer is given in an alternative form:$\displaystyle 6\times{^5C_2}\div\frac{9!}{4!3!2!}$

The reason is that, with 6 available 'slots' (1 slot for the block of 4 I's and 5 slots for the remaining 5 letters), there are 6 ways of selecting the position of the block of 4 I's, and then $\displaystyle {^5C_2}$ ways of choosing the two slots to be occupied by the 2 P's; the 3 S's filling the remaining 3 slots.

3) Use either of these two techniques to answer this question, where (once the 4 S's have been placed at the end) there are 7 items left with 4 alike of the first kind (I) and 2 alike of the second kind (P).

Can you complete this one now?

Grandad