Question 1:
A bag contains 36 marbles which are black and white. It is given that the probability for a black marble being picked at random from the bag is $\frac{5}{9}$ .
Calculate the number of white marbles to be taken out from the bag so that the probability of picking a black marble is $\frac{5}{8}$ .
Solution:
$\begin{array}{l}\text{Numberofblackmarblesinthebag}\\ \text{=}\frac{5}{9}\times 36=20\\ \\ \text{Let}y\text{isthetotalnumberofmarblesleftinthebag}\text{.}\\ y\times \frac{5}{8}=20\\ y=20\times \frac{8}{5}=32\\ \\ \text{Numberofwhitemarblestobetakenoutfromthebag}\\ \text{=}3632\\ =4\end{array}$
A bag contains 36 marbles which are black and white. It is given that the probability for a black marble being picked at random from the bag is $\frac{5}{9}$ .
Calculate the number of white marbles to be taken out from the bag so that the probability of picking a black marble is $\frac{5}{8}$ .
Solution:
$\begin{array}{l}\text{Numberofblackmarblesinthebag}\\ \text{=}\frac{5}{9}\times 36=20\\ \\ \text{Let}y\text{isthetotalnumberofmarblesleftinthebag}\text{.}\\ y\times \frac{5}{8}=20\\ y=20\times \frac{8}{5}=32\\ \\ \text{Numberofwhitemarblestobetakenoutfromthebag}\\ \text{=}3632\\ =4\end{array}$
Question 2:
Solution:
Probability of picking a bag = $\frac{1}{3}$
Probability of picking purple ball from bag A = $\frac{6}{10}=\frac{3}{5}$
Probability of picking purple ball from bag B = $\frac{4}{12}=\frac{1}{3}$
Probability of picking purple ball from bag C = $\frac{2}{12}=\frac{1}{6}$
$\begin{array}{l}P\left(\text{purple ball}\right)\text{=}\left(\frac{1}{3}\times \frac{3}{5}\right)+\left(\frac{1}{3}\times \frac{1}{3}\right)+\left(\frac{1}{3}\times \frac{1}{6}\right)\\ \text{}=\frac{1}{5}+\frac{1}{9}+\frac{1}{18}\\ \text{}=\frac{11}{30}\end{array}$
Table below shows the number of different coloured balls in three bags.
Green 
Brown 
Purple 

Bag A 
3 
1 
6 
Bag B 
5 
3 
4 
Bag C 
4 
6 
2 
If a bag is picked at random and then a ball is drawn randomly from that bag, what is the probability that a purple ball is drawn?
Solution:
Probability of picking a bag = $\frac{1}{3}$
Probability of picking purple ball from bag A = $\frac{6}{10}=\frac{3}{5}$
Probability of picking purple ball from bag B = $\frac{4}{12}=\frac{1}{3}$
Probability of picking purple ball from bag C = $\frac{2}{12}=\frac{1}{6}$
$\begin{array}{l}P\left(\text{purple ball}\right)\text{=}\left(\frac{1}{3}\times \frac{3}{5}\right)+\left(\frac{1}{3}\times \frac{1}{3}\right)+\left(\frac{1}{3}\times \frac{1}{6}\right)\\ \text{}=\frac{1}{5}+\frac{1}{9}+\frac{1}{18}\\ \text{}=\frac{11}{30}\end{array}$