Answer to Question #153239 in Statistics and Probability for Sadisha

Question

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

The bag contains 6 blue balls, 5 green balls and 4 red balls.

$\therefore$ Total number of balls is 15.

a)

1st Selection :-

Probability of selecting $1$ of the 6 blue balls out of 15 balls $= \frac{6 \choose 1}{15\choose1}=\frac{6}{15}$

After selecting the 1st ball, 5 blue balls, 5 green balls and 4 red balls are remaining in the bag.

[ As replacement is not allowed ]

$\therefore$ Total number of remaining balls is 14.

2nd Selection :-

Probability of selecting $1$ of the 5 blue balls out of 14 balls $= \frac{5 \choose 1}{14\choose1}=\frac{5}{14}$

After selecting the 2nd ball, 4 blue balls, 5 green balls and 4 red balls are remaining in the bag.

[ As replacement is not allowed ]

$\therefore$ Total number of remaining balls is 13.

3nd Selection :-

Probability of selecting $1$ of the 4 blue balls out of 13 balls $= \frac{4 \choose 1}{13\choose1}=\frac{4}{13}$

So, Probability of occurring above 3 events in a row $=\frac{6}{15}\times \frac{5}{14}\times \frac{4}{13}=\frac{4}{91}$

$\therefore$ Probability of the event that all selected balls are blue $=\frac{4}{91}$ (Ans.)

b)

1st Selection :-

Probability of selecting $1$ of the 6 blue balls out of 15 balls $= \frac{6 \choose 1}{15\choose1}=\frac{6}{15}$

After selecting the 1st ball, 5 blue balls, 5 green balls and 4 red balls are remaining in the bag.

[ As replacement is not allowed ]

$\therefore$ Total number of remaining balls is 14.

2nd Selection :-

Probability of selecting $1$ of the 5 blue balls out of 14 balls $= \frac{5 \choose 1}{14\choose1}=\frac{5}{14}$

After selecting the 2nd ball, 4 blue balls, 5 green balls and 4 red balls are remaining in the bag.

[ As replacement is not allowed ]

$\therefore$ Total number of remaining balls is 13.

3nd Selection :-

Probability of selecting $1$ of the 5 green balls out of 13 balls $= \frac{5 \choose 1}{13\choose1}=\frac{5}{13}$

So, Probability of occurring above 3 events in a row $=\frac{6}{15}\times \frac{5}{14}\times \frac{5}{13}=\frac{5}{91}$

$\therefore$ Probability of the event that selected two balls are blue and one is green $=\frac{5}{91}$ (Ans.)

c)

1st Selection :-

Probability of selecting $1$ of the 6 blue balls out of 15 balls $= \frac{6 \choose 1}{15\choose1}=\frac{6}{15}$

After selecting the 1st ball, 5 blue balls, 5 green balls and 4 red balls are remaining in the bag.

[ As replacement is not allowed ]

$\therefore$ Total number of remaining balls is 14.

2nd Selection :-

Probability of selecting $1$ of the 5 green balls out of 14 balls $= \frac{5 \choose 1}{14\choose1}=\frac{5}{14}$

After selecting the 2nd ball, 5 blue balls, 4 green balls and 4 red balls are remaining in the bag.

[ As replacement is not allowed ]

$\therefore$ Total number of remaining balls is 13.

3nd Selection :-

Probability of selecting $1$ of the 4 red balls out of 13 balls $= \frac{4 \choose 1}{13\choose1}=\frac{4}{13}$

So, Probability of occurring above 3 events in a row $=\frac{6}{15}\times \frac{5}{14}\times \frac{4}{13}=\frac{4}{91}$

$\therefore$ Probability of the event that there is one of each colour $=\frac{4}{91}$ (Ans.)