Question

A box contains $10$ red marbles, $20$ blue marbles and $30$ green marbles. $5$ marbles are drawn from the box, what is the probability that

$\left(i\right)$ all will be blue$?$

$\left(ii\right)$ At least one will be green$?$

Answer 1

$\left(i\right)$ Required probability
Given : A box contains $60$ marbles$(10$ red, $20$ blue & $30$ green$).$

We need to chhose $5$ marbles out of $60$ marbles. Let sample space $S$

Let $A$ be the event that all marbles are blue

$n\left(S\right)={}^{60}{C}_5$ and $n\left(A\right)={}^{20}{C}_5$

$p$$($all marbles are blue$)$ = $\frac{\text{favourable outcomes}}{\text{Total out comes}}$

$\Rightarrow P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{{}^{20}{C}_5}{{}^{60}{C}_5}$

Hence the required probability is $\frac{{}^{20}{C}_5}{{}^{60}{C}_5}$

$\left(ii\right)$ Required probability
Given : A box contains $60$ marbles $(10$ red, $20$ blue & $30$ green$).$

We need to chhose $5$ marbles out of $60$ marbles. Let sample space $S$

Let $B$ be the event that none of marbles are green $($i.e., All are chosen from red and blue$)$

$n\left(S\right)={}^{60}{C}_5$ and $n\left(B\right)={}^{30}{C}_5$

$p$$($at least one will be green$)$ = $1-P$( none of marbles are green $)$

$P\left(B\right)=1-\frac{n\left(B\right)}{n\left(S\right)}=1-\frac{{}^{30}{C}_5}{{}^{60}{C}_5}$

Hence the required probability is $1-\frac{{}^{30}{C}_5}{{}^{60}{C}_5}$.