Question

A bag contains 5 red balls and 8 blue balls. It also contains 4 green and 7 black balls if a ball is drawn at random, then find the probability that it is not green.
A) $\dfrac{5}{6}$
B) $\dfrac{1}{4}$
C) $\dfrac{1}{6}$
D) $\dfrac{7}{4}$

Answer 1

Hint:
Total balls is equal to 24 balls. Total no of ways of selecting one ball is ${}^{24}{C_1} = 24$. probability of not selecting green ball is equal to sum of probability of selecting red blue and black ball. Probability of selecting red ball is $\dfrac{{{}^5{C_1}}}{{{}^{24}{C_1}}}$ and that of blue and black ball is$\dfrac{{{}^8{C_1}}}{{{}^{24}{C_1}}}$ and $\dfrac{{{}^7{C_1}}}{{{}^{24}{C_1}}}$ respectively.

Complete step by step solution:
Given, a bag contains 5 red balls, 8 blue balls, 4 green and 7 black balls
Total no. of ways to draw a ball is ${}^{24}{C_1} = 24$
No. of ways of selecting red balls =${}^5{C_1} = 5$
No. of ways of selecting blue balls =${}^8{C_1} = 8$
No. of ways of selecting black balls =\[{}^7{C_1} = 7\]
Now, probability of not selecting green ball = sum of probability of selecting red blue and black ball
$ = \dfrac{{{}^5{C_1}}}{{{}^{24}{C_1}}} + \dfrac{{{}^8{C_1}}}{{{}^{24}{C_1}}} + \dfrac{{{}^7{C_1}}}{{{}^{24}{C_1}}}$
$ = \dfrac{{5 + 8 + 7}}{{24}}$
$ = \dfrac{5}{6}$

Hence, Option A is the correct option.

Note:
Alternative method
\[Probability{\text{ }}of{\text{ }}not{\text{ }}selecting{\text{ }}green{\text{ }} = 1 - probability{\text{ }}of{\text{ }}selecting{\text{ }}green\]
Probability of selecting green$ = \dfrac{{{}^4{C_1}}}{{{}^{24}{C_1}}} = \dfrac{1}{6}$
Probability of not selecting green$ = 1 - \dfrac{1}{6} = \dfrac{5}{6}$