Question

If there are 5 white balls 3 black balls and 2 red balls in a bag, then what is the probabibilty that if we pick 3 balls they are of different colors?

Answer 1

We want the number of ways to choose a white ball, multiplied by the number of ways to choose a black ball, multiplied by the number of ways to choose a red ball.....and we divide this by the total number of ways to pick 3 balls from the 10.

So we have

\[\frac{ \left(\begin{matrix}4 \\ 1\end{matrix}\right)\left(\begin{matrix}4 \\ 1\end{matrix}\right)\left(\begin{matrix}2 \\ 1\end{matrix}\right) }{ \left(\begin{matrix}10 \\ 3\end{matrix}\right)}\]

\[=\frac{ 4*4*2 }{ \left(\begin{matrix}10 \\ 3\end{matrix}\right) }\]

\[=\frac{ 32 }{ \left(\begin{matrix}10 \\ 3\end{matrix}\right) }\]

=12/45 I believe (did the math in my head really fast, double check this)

Answer 2

are there choices? i got 2/13

Answer 3

Actually i dont have the answer! :/

Answer 4

I have changed the ques!
Now the answer is 1/4!

Answer 5

@GoBlue now plz calculate
I have changd the ques!

Answer 6

on similar lines,
\(\huge\frac{ \left(\begin{matrix}5 \\ 1\end{matrix}\right)\left(\begin{matrix}3 \\ 1\end{matrix}\right)\left(\begin{matrix}2 \\ 1\end{matrix}\right) }{ \left(\begin{matrix}10 \\ 3\end{matrix}\right)}=\frac{1}{4}\)

Answer 7

yes! @harton rite! :)