0
$\begingroup$

Each of the letters in the word "MATHEMATICS" is on a letter tile in a bag. Foool picks three without replacement. what is the probability that he will get all vowels?

My approach,

The number of ways to the letters of the words "MATHEMATICS" can be arranged 3 at a time is coefficient of $x^3 $ $$ 3! \times (1+x)^5 \times \left(1+x+\frac{x^2}{2}\right)^3$$ which is $399$

The number of arrangements of 2A's, 1I and 1E taken 3 at at a time is coefficient of $x^3$ in $$ 3! \times (1+x)^2 \times \left(1+x+\frac{x^2}{2}\right)$$ which is $12$.

Then the required probability is given by $\frac {12}{399} $ but apparently this is not the right answer. What exactly I am missing here?

$\endgroup$

3 Answers 3

Reset to default
7
$\begingroup$

How many ways are there to pick 3 letters? How many ways are there to pick 3 vowels? By my count, the right answer should be $$\binom{4}{3}/\binom{11}{3}=\frac{4}{165}.$$

$\endgroup$
3
  • $\begingroup$ Thanks Arturo, I just over-complicated a simple problem :( $\endgroup$
    – Quixotic
    Commented Mar 4, 2012 at 8:36
  • 1
    $\begingroup$ You have four items: four physical tiles; they are different objects, because they are different tiles. Just because two of them have the same letter written on them does not make them the same tile! $\endgroup$
    – Arturo Magidin
    Commented Mar 4, 2012 at 8:40
  • $\begingroup$ Thanks again Arturo, I was all messed up. $\endgroup$
    – Quixotic
    Commented Mar 4, 2012 at 8:42
1
$\begingroup$

I don't know what you're missing, because I don't know why you're taking the approach you take. The answer is $(4/11)(3/10)(2/9)$.

$\endgroup$
4
  • $\begingroup$ But there is repetition of letters, so why would this work? $\endgroup$
    – Quixotic
    Commented Mar 4, 2012 at 8:30
  • 1
    $\begingroup$ What difference does repetition make? If it were MATHEMOTICS, how would it be different? $\endgroup$
    – Gerry Myerson
    Commented Mar 4, 2012 at 8:32
  • $\begingroup$ Why do repetitions matter? You have one tile for each letter, so you have 11 tiles. Of these 11 tiles, 4 tiles contain vowels. If you grab 3 tiles, what are the odds that all 3 tiles are vowels? Where the letters come from, or what the letters are, is irrelevant; the problem works exactly the same way if four tiles have an A in them and seven tiles have a B; or if four tiles are A, E, I, O, and the other seven are B, C, D, F, G, H, J. You are not being asked for the possible number of 3-letter sequences that can be made from the letters in "MATHEMATICS" (where repetitions would matter). $\endgroup$
    – Arturo Magidin
    Commented Mar 4, 2012 at 8:32
  • $\begingroup$ Foool: You are confusing this question with the question "What are the different possibilities of picking three vowels?". In this case the repetitions would matter. $\endgroup$
    – fretty
    Commented Mar 4, 2012 at 9:12
-1
$\begingroup$

We obtained vowels in a word =5 (a ,e i, o ,u ) And numbers of word = 11

A/Q the probability of picking up a vowel = 5/ 11 ans

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .