**Question 2:**

A bag contains 4 red cards, 6
blue cards and 5 green cards. A card is drawn at random and its colour is
recorded and then it is returned to the bag before the second card is drawn at
random. Find the probability that

(a) all the three cards are
blue,

(b) there are two blue cards
followed by one red card,

(c) the sequence of the cards
drawn is red, green and blue,

(d) all the three cards have
the same colour.

*Solution:*
Let

*R*= red card,*B*= blue card,*G*= green card**(a)**

**(b)**

$\begin{array}{l}\text{Probability(twobluecardsfollowedbyoneredcard})\\ =\frac{6}{15}\times \frac{6}{15}\times \frac{4}{15}\\ =\frac{16}{375}\end{array}$

**(c)**

$\begin{array}{l}\text{Probability(thesequenceofthecardsdrawnisred,greenandblue})\\ =\frac{4}{15}\times \frac{6}{15}\times \frac{5}{15}\\ =\frac{8}{225}\end{array}$

**(d)**

$\begin{array}{l}\text{Probability(allthethreecardshavethesamecolour})\\ =P\left(RRR\right)+P\left(BBB\right)+P\left(GGG\right)\\ ={\left(\frac{4}{15}\right)}^{3}+{\left(\frac{6}{15}\right)}^{3}+{\left(\frac{5}{15}\right)}^{3}\\ =\frac{64}{3375}+\frac{216}{3375}+\frac{125}{3375}\\ =\frac{3}{25}\end{array}$

**Question 3:**

A bag contains 4 blue beads,
3 red beads and 7 green beads. Two beads are drawn at random from the bag, one
after the other without replacement. Find the probability that

(a) both the beads are of the
same colour,

(b) both the beads are of
different colours.

*Solution:***(a)**

$\begin{array}{l}\text{Probability(}both\text{}the\text{}beads\text{}are\text{}of\text{}the\text{}same\text{}colour)\\ =P\left(\text{blue,blue}\right)+P\left(\text{red,red}\right)+P\left(\text{green,green}\right)\\ =\left(\frac{4}{14}\times \frac{3}{13}\right)+\left(\frac{3}{14}\times \frac{2}{13}\right)+\left(\frac{7}{14}\times \frac{6}{13}\right)\\ =\frac{6}{91}+\frac{3}{91}+\frac{3}{13}\\ =\frac{30}{91}\end{array}$

**(b)**

$\begin{array}{l}\text{Probability(boththebeadsareofdifferentcolour})\\ =1-P\left(\text{boththebeadsareofthesamecolour}\right)\\ =1-\frac{30}{91}\leftarrow \overline{)\text{Frompart(a)}}\\ =\frac{61}{91}\end{array}$