Probability Questions - Answers don't match with the textbook

[Ossih]

Hi! I was doing some questions from the IBID Textbook on Probability, and my answers don't match the answers at the back. Could someone help explain these for me?

1. (a) Write down the number of different arrangements of the letters of the word EQUILIBRIUM. ( i got that one right)

(b) One of these arrangements is chosen at random. Find the probability that:

(i) the first two letters are consonants (the answer at the back is 2/11)

(ii) all the vowels are together (the answer at the back is 2/77)

2. In a 3 match test series between NZ and Australia, New Zealand has the probability of winning 0.2 per match, 0.3 of a draw, and Australia has a probability of 0.5 per match. Find the probability that Australia wins the series, that is, wins more matches than New Zealand. (the answer at the back is 0.635)

My Solutions

For number 1(a), I just did 11! / (2!3!)

For (b)(i) I did (6 P 2)(9!)/(11!)

For (b)(ii) I treated all the vowels as a single letter, so (7!)(5!)/(11!)

For number 2, I drew a tree diagram and included the cases:

(NZ - AU - AU) + (D - AU - AU) + (AU - NZ - AU) + (AU - D - AU) + (AU - AU - AU) + (AU - AU - D) + (AU - AU - NZ)

= (0.2)(0.5)(0.5) + (0.3)(0.5)(0.5) + (0.5)(0.2)(0.5) + (0.5)(0.3)(0.5) + (0.5)(0.5)(0.5) + (0.5)(0.5)(0.3) + (0.5)(0.5)(0.2)

Thank you so much!

Edited May 16, 2014 by Ossih

[trmesd2]

Assume U is vowel. We have 5 consonants: QLBRM and 6 vowels: EUUIII

For (b)(i)

I got the same answer as the solution: position1,2: 5x4 combinations, positions 3-11 combinations: 9!

Therefore all combinations: 5x4x9! But we also have all combinations=11!, therefore probability for (b)(i) is 5x4x9!/11!=2/11 exactly the same as the given anwser

For (b)(ii) I got a different answer:

Imagine we have a block of (EUUIII) that cannot be sepatated, noted here as block. Then we have 6 elements to arrange: block,QLBRM, so the no. of combinations is 6! But for each block, we have 6! combinations(we need to count 2 Us and 3 IIIs as unique because on the surface they look the same but they exert the same chances of appearing), therefore the total no. of combinations for (b)(ii) is 6! x 6! / 11! = 1/77. I consider the given answer is wrong!