Question

A bag contains two red balls, three blue balls and five green balls.
Three balls are drawn at random. Find the probability that
a) the three balls are of different colours
b) two balls are of the same colour
c) all the three are of the same colour.

Answer 1

Answer on Question #44494 – Math – Statistics and Probability

A bag contains two red balls, three blue balls and five green balls. Three balls are drawn at random. Find the probability that:

a) three balls are of different colours;

b) two balls are of the same colour;

c) all the three are of the same colour.

Solution.

a)

$$P(\text{three balls are of different colours}) = \frac{M}{N},$$

$M$ is the number of ways to select 1 red ball, 1 blue ball and 1 green ball,

$N$ is the number of ways to select 3 balls from the set of 10 balls;

So:

$$M = \binom{2}{1} \binom{3}{1} \binom{5}{1} = 2 \cdot 3 \cdot 5 = 30;$$ $$N = \binom{10}{3} = \frac{10!}{3! \cdot 7!} = \frac{8 \cdot 9 \cdot 10}{6} = 8 \cdot 3 \cdot 5 = 120;$$ $$P(\text{three balls are of different colours}) = \frac{30}{120} = 0.25.$$

b)

$$\begin{array}{l} P(\text{two balls are of the same colour}) = \\ = P(2 \text{ balls are red and } 1 \text{ isn't red}) + P(2 \text{ balls are blue and } 1 \text{ isn't blue}) + \\ + P(2 \text{ balls are green and } 1 \text{ isn't green}) = \frac{P + Q + R}{N}, \end{array}$$

$P$ is the number to select 2 red balls and 1 ball which is not red,

$Q$ is the number to select 2 blue balls and 1 ball which is not blue,

$R$ is the number to select 2 green balls and 1 ball which is not green,

$N$ is the same as in (a);

So:

$$P = \binom{2}{2} \binom{8}{1} = 1 \cdot 8 = 8;$$ $$Q = \binom{3}{2} \binom{7}{1} = 3 \cdot 7 = 21;$$ $$R = \binom{5}{2} \binom{5}{1} = \frac{5!}{2! \cdot 3!} \cdot 5 = 10 \cdot 5 = 50;$$ $$P(\text{two balls are of the same colour}) = \frac{8 + 21 + 50}{120} = \frac{79}{120}.$$

c)

Note that $P(3 \text{ balls are red}) = 0$.

$$\begin{array}{l} P(\text{all the three are of the same colour}) = \\ = P(3 \text{ balls are red}) + P(3 \text{ balls are blue}) + P(3 \text{ balls are green}) = 0 + \frac{S + T}{N}, \end{array}$$

$S$ is the number of ways to select 3 blue balls,

$T$ is the number of ways to select 3 green balls,

$N$ is the same as in (a);

So:

$$S = \binom{3}{3} = 1;$$ $$T = \binom{5}{3} = \frac{5!}{3! \cdot 2!} = 10;$$

$P(\text{all the three are of the same colour}) = \frac{1 + 10}{120} = \frac{11}{120}$.

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