Question

5. A bag contains 6 blue balls, 5 green balls and 4 red balls. Three are selected at random without replacement. Find the probability that
(a) they are all blue
(b)two are blue and one is green
(c) there is one of each colour

Answer 1

Total balls $=6+5+4=15$

$B=6, G=5, R=4$

a. Probability three blue balls are picked without replacement is given by $P(B)\times{P(B)}\times{P(B)}$

$=\frac{6}{15}\times{\frac{5}{14}}\times{\frac{4}{13}}=\frac{120}{2730}=\frac{4}{91}$

b. Probability that two blue balls and one green ball are picked is given by $[P(B)\times{P(B)}\times{P(G)}]+[P(B)\times{P(G)}\times{P(B)}]+[P(G)\times{P(B)}\times{P(B)}]$

$=[\frac{6}{15}\times{\frac{5}{14}}\times{\frac{5}{13}}]+[\frac{6}{15}\times{\frac{5}{14}}\times{\frac{5}{13}}]+[\frac{5}{15}\times{\frac{6}{14}}\times{\frac{5}{13}}]$

$=3\times{\frac{150}{2730}}=\frac{15}{91}$

c. Probability that 1 blue, 1 green and 1 red balls are picked is given by

$[P(B)\times{P(G)}\times{P(R)}]+[P(B)\times{P(R)}\times{P(G)}]+[P(R)\times{P(B)}\times{P(G)}]$

$+[P(R)\times{P(G)}\times{P(B)}]+[P(G)\times{P(R)}\times{P(B)}]+[P(G)\times{P(B)}\times{P(R)}]$

$=[\frac{6}{15}\times{\frac{5}{14}}\times{\frac{4}{13}}]+[\frac{6}{15}\times{\frac{4}{14}}\times{\frac{5}{13}}]+[\frac{4}{15}\times{\frac{6}{14}}\times{\frac{5}{13}}]+$

$[\frac{4}{15}\times{\frac{5}{14}}\times{\frac{6}{13}}]+[\frac{5}{15}\times{\frac{4}{14}}\times{\frac{6}{13}}]+[\frac{5}{15}\times{\frac{6}{14}}\times{\frac{4}{13}}]$

$=6\times{\frac{120}{2730}}=\frac{24}{91}$