Question

A bag contains 30 coloured balls. Of these, 10 are red, 6 are blue and the
rest are green or yellow.
A ball is taken at random from the bag. The probability that this ball is
yellow is 1/6
.
What is the probability that a ball taken at random from the bag is:
(a) green,
(b) green or yellow,
(c) red or blue,
(d) red, blue or yellow?

Answer 1

Dear student,

$$\begin{gathered} Totalno.ofballs=30 \\ No.ofredballs=10 \\ ProbabilityofredballsP\left(R\right)=\frac{No.ofredballs}{Totalno.ofballs}=\frac{10}{30}=\frac{1}{3} \\ No.ofblueballs=6 \\ ProbabilityofblueballsP\left(B\right)=\frac{No.ofblueballs}{Totalno.ofballs}=\frac{6}{30}=\frac{1}{5} \\ ProbabilityofyellowballsP\left(Y\right)=\frac{No.ofyellowballs}{Totalno.ofballs}=\frac{1}{6} \\ No.ofgreenoryellowballs=30-10-6=14 \\ Probabilityofgreenoryellowballs=\frac{No.ofgreenoryellowballs}{Totalno.ofballs}=\frac{14}{30} \\ Probabilityofgreenoryellowballs=Probabilityofgreen+Probabilityofyellow \\ \Rightarrow \frac{14}{30}=P\left(G\right)+\frac{1}{6} \\ \Rightarrow \frac{14}{30}-\frac{1}{6}=P\left(G\right) \\ \Rightarrow P\left(G\right)=\frac{14-5}{30} \\ \Rightarrow P\left(G\right)=\frac{9}{30} \\ a)Probabilitythataballtakenatrandomfromthebagisgreen=\frac{9}{30}=\frac{3}{10} \\ b)Probabilitythataballtakenatrandomfromthebagisgreenoryellow=\frac{14}{30} \\ c)Probabilitythataballtakenatrandomfromthebagisredorblue=P(R)+P(B) \\ =\frac{1}{3}+\frac{1}{5} \\ =\frac{5+3}{15} \\ =\frac{8}{15} \\ d)Probabilitythataballtakenatrandomfromthebagisred,blueoryellow=P\left(R\right)+P\left(B\right)+P\left(Y\right) \\ =\frac{1}{3}+\frac{1}{5}+\frac{1}{6} \\ =\frac{10+6+5}{30} \\ =\frac{2}{30} \end{gathered}$$
Regards