Question

In: Math

A random rearrangement doe not separate out between repeated letters. Consider word CHRISTMASTIME. a) What's the...

A random rearrangement doe not separate out between repeated letters. Consider word CHRISTMASTIME.

a) What's the expected # vowels in first three letters of random rearrangement of CHRISTMASTIME?

b) What's probability that all the S's happen before all the I's in random rearrangement of CHRISTMASTIME?

c) What's probability that word CHRIST happens in consecutive letters of uniform rearrangement of CHRISTMASTIME?

Solutions

Expert Solution

a) Total number of letters: 13

Total number of vowels: 4

Therefore the expected number of vowels in the first 3 letters of a random rearrangement here is computed as:

= (4/13)*3 = 0.9231

Therefore 0.9231 is the expected number of vowels here

b) The relative position of all the S and I are analyzed here.

We have 2 Is and 2 Ss here.

Thus total number of relative arrangements of these 4 letters is given as:
= Total number of permutation of 4 letters given that there are 2 pair of similar letters

= 4! / (2*2)

= 6

Total positions where all S comes before all Is

= 1 which is the form SSII

Therefore the required probability here is given as: 1/6

Therefore 1/6 = 0.1667 is the required probability here.

c) Let us keep CHRIST as one group and the rest of the letters as separate letters which are:
M - 2, A, S, T, I, E

Therefore total permutations here is computed as:
= 7! / 2

= 2520

Also total number of permutations without any condition here is computed as:
Total letters: C, H, R, I-2, S-2, T-2, M-2, A, E
= 13! / (24)
= 389188800

Therefore the required probability here is computed as:
= 2520/389188800

= 0.000006475006475

Therefore 0.000006 is the required probability here.


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