# Thread: probability of choosing two word

1. ## probability of choosing two word

Hi

Can anybody tell wether i am right or not.

Q. Two WORDS are chosen from word "SUMMER", what is probability that at least M is there?

Total Nos of chosen word are : 6C2=15
SU SM SM SE SR
UM UM UE UR
MM ME MR
ME MR
ER

SO P(at least one is M)=9/15

Something seemed odd so I used to the hypergeometric distribution formula to make sure. 9/15 is the correct answer

$\displaystyle \frac{{{2}\choose{1}}{{4}\choose{1}}}{{{6}\choose{ 2}}}$$\displaystyle +\frac{{{2}\choose{2}}{{4}\choose{0}}}{{{6}\choose {2}}}=\frac{9}{15}$

3. Thank you...

4. Hello, amb03!

$\displaystyle \text{Can anybody tell whether i am right or not.?}$

$\displaystyle \text{A two-letter "word" is chosen from word SUMMER.}$
$\displaystyle \text{What is probability that at least one M is chosen?}$

$\displaystyle \text{Total number of chosen words is: }\;_6C_2\:=\:15$

. . $\displaystyle \begin{array}{c} \text{SU SM SM SE SR} \\ \text{UM UM UE UR} \\ \text{MM ME MR} \\ \text{ME MR} \\ \text{ER} \end{array}$ .Good!

$\displaystyle \text{So: }\;P(\text{at least one M}) \:=\:\dfrac{9}{15}$ . Correct!

I solved it from "the other end" . . .

How many words have no M's?

There are 2 M's and 4 Others.

We want 2 of the Others.
. . There are: .$\displaystyle _4C_2 \:=\:6$ words with no M's.

Hence, there are: .$\displaystyle 15 - 6 \:=\:9$ words with at least one M.

Therefore: .$\displaystyle P(\text{at least one M}) \:=\:\dfrac{9}{15}$