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 = $\frac{2}{12}=\frac{1}{6}$
$\begin{array}{l}P\left(\text{purple ball}\right)\text{=}\left(\frac{1}{3}\times \frac{3}{5}\right)+\left(\frac{1}{3}\times \frac{1}{3}\right)+\left(\frac{1}{3}\times \frac{1}{6}\right)\\ \text{}=\frac{1}{5}+\frac{1}{9}+\frac{1}{18}\\ \text{}=\frac{11}{30}\end{array}$
Question 3:
A box contains 48 marbles. There are red marbles and green marbles. A marble is chosen at random from the box. The probability that a red marble is chosen is $\frac{1}{6}.$
How many red marbles need to be added to the box so that the probability that a red marble is chosen is $\frac{1}{2}.$
Solution:
$\begin{array}{l}\text{Numberofredmarblesinthebox}\\ \text{=}\frac{1}{6}\times 48\\ =8\\ \\ \text{Letthenumberofredmarblesneededtobeaddedbe}y\text{.}\\ \text{P}\left(\text{redmarble}\right)=\frac{1}{2}\\ \frac{8+y}{48+y}=\frac{1}{2}\\ 16+2y=48+y\\ 2yy=4816\\ y=32\\ \\ \therefore \text{Numberofredmarblesneedtobeadded}=32\end{array}$
A box contains 48 marbles. There are red marbles and green marbles. A marble is chosen at random from the box. The probability that a red marble is chosen is $\frac{1}{6}.$
How many red marbles need to be added to the box so that the probability that a red marble is chosen is $\frac{1}{2}.$
Solution:
$\begin{array}{l}\text{Numberofredmarblesinthebox}\\ \text{=}\frac{1}{6}\times 48\\ =8\\ \\ \text{Letthenumberofredmarblesneededtobeaddedbe}y\text{.}\\ \text{P}\left(\text{redmarble}\right)=\frac{1}{2}\\ \frac{8+y}{48+y}=\frac{1}{2}\\ 16+2y=48+y\\ 2yy=4816\\ y=32\\ \\ \therefore \text{Numberofredmarblesneedtobeadded}=32\end{array}$
Question 4:
A box contains 5 red cards, 3 yellow cards and a number of green cards. A card is picked at random from the box. Given that the probability of picking a yellow card is $\frac{1}{6}$ , find the probability of picking a card that is not green.
Solution:
$\begin{array}{l}\text{P}\left(\text{yellowcard}\right)=\frac{n\left(\text{yellowcard}\right)}{n\left(S\right)}\\ \text{}\frac{1}{6}=\frac{3}{n\left(S\right)}\\ \text{}n\left(S\right)=3\times 6\\ \text{}=18\\ \\ n\left(\text{notgreencard}\right)=5+3=8\\ \text{P}\left(\text{notgreencard}\right)=\frac{8}{18}\\ \text{}=\frac{4}{9}\end{array}$
A box contains 5 red cards, 3 yellow cards and a number of green cards. A card is picked at random from the box. Given that the probability of picking a yellow card is $\frac{1}{6}$ , find the probability of picking a card that is not green.
Solution:
$\begin{array}{l}\text{P}\left(\text{yellowcard}\right)=\frac{n\left(\text{yellowcard}\right)}{n\left(S\right)}\\ \text{}\frac{1}{6}=\frac{3}{n\left(S\right)}\\ \text{}n\left(S\right)=3\times 6\\ \text{}=18\\ \\ n\left(\text{notgreencard}\right)=5+3=8\\ \text{P}\left(\text{notgreencard}\right)=\frac{8}{18}\\ \text{}=\frac{4}{9}\end{array}$