# Probability of picking two letters from the word MATHEMATICAL [closed]

If you select 3 letters (with replacement) from the word MATHEMATICAL, what is the probability of getting two 'M's.

-

## closed as off-topic by mathlove, PVAL, DeepSea, Claude Leibovici, Tunk-FeyAug 25 '14 at 11:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – mathlove, PVAL, DeepSea, Claude Leibovici, Tunk-Fey
If this question can be reworded to fit the rules in the help center, please edit the question.

Thanks for the answers. I get 5/72 as well. But the correct answer is 11/144. Could some one please explain this answer to me. Thx. – Susie Aug 25 '14 at 11:08
I do not agree with the answer of 11 / 144. – bobbym Aug 25 '14 at 11:12
I tend to agree with you. But the answer in the text book is 11/144 – Susie Aug 25 '14 at 11:16
The textbook's answer is wrong. – David Mitra Aug 25 '14 at 11:21
Much appreciated. Thank you. – Susie Aug 25 '14 at 11:24

How many letters are there in the word MATHEMATICAL? $12$

How many letters are $M$ here? $2$

We want the probability of getting two 'M's takes considering all three cases. When we get a non-M on the first, second or third pick.

$\frac{10}{12}*\frac{2}{12}*\frac{2}{12}+\frac{2}{12}*\frac{10}{12}*\frac{2}{12}+\frac{2}{12}*\frac{2}{12}*\frac{10}{12}=30*\frac{2^2}{12^3}=\frac{5}{72}$

-

The number of M's obtained is binomially distributed with $p={1\over6}$. The probability to obtain exactly two M's in three draws is therefore given by $${3\choose 2}\ \left({1\over6}\right)^2\ \left({5\over6}\right)^1={5\over72}\ .$$

-

I am getting,

$$3\left ( \frac{2}{12}\right )\left ( \frac{2}{12}\right )\left ( \frac{10}{12}\right )=\frac{5}{72}$$

-
Why do we have three answers? What is wrong with my method and/or $8\pi \mathrm{r}$'s? – Tony Aug 25 '14 at 11:05
I think because I am treating (m,m,X) as different than (X,m,m) and (m,X,m). – bobbym Aug 25 '14 at 11:09
Oh good point, thank you for that! – Tony Aug 25 '14 at 11:10
I am your upvote here, and I have no idea why I got one more upvote than you, considering my answer was obtained from your help! – Tony Aug 26 '14 at 14:42
It us not a big deal, glad you got them. Looks like they are closing the question anyway. – bobbym Aug 26 '14 at 15:08

Number of 'M's in MATHEMATICAL = 2

Number of letters in MATHEMATICAL = 12

Probability of getting an 'M': $\frac{2}{12}$

Probability of not getting an 'M': $\frac{10}{12}$

Ways to shuffle these three: $\binom{3}{2} = 3$ ($3$ letters of which $2$ are identical)

Probability: $$\left ( \frac{2}{12} \right )^2 \cdot \frac{10}{12} \cdot \binom{3}{2} = \frac{5}{72}$$

-