1. ## Question on probability

Hi, I have a question on probability here,

Five letters are chosen from the word "singapura", find the probability that the letters chosen can be used to form the word "singa".

is it 2/(9C5) = 1/63?

Is it possible to solve this question without using combination (nCr)?

2. Hello, acc100jt!

Five letters are chosen from the word SINGAPURA.
Find the probability that the letters chosen can be used to form the word SINGA.

is it: . $\frac{2}{_9C_5} \:=\:\frac{1}{63}$ ? . . . . Yes!
Is it possible to solve this question without using combination (nCr)?
It is possible, but it takes more work . . .

(1) We can list the 126 possible 5-letter samples
. . . and then count the number that contain S,I,N,G,A (there are two).

(2) We could work out the probability like this.

Find the probability of getting S-I-N-G-A (in that order).

$\text{We have: }\;\underbrace{\frac{1}{9}}_S \cdot\underbrace{\frac{1}{8}}_I \cdot\underbrace{\frac{1}{7}}_N \cdot\underbrace{\frac{1}{6}}_G\cdot\underbrace{\f rac{2}{5}}_A \;=\;\frac{1}{7560}$

Since the order is unimportant, the probability is: . $5! \times\frac{1}{7560} \;=\;\frac{1}{63}$

3. Yes, you can.

(1/9)(1/8)(1/7)(1/6)(2/5)=1/7560

But they can be arranged in 5!=120 ways, so

$120(1/7560)=\frac{1}{63}$