Question

A bag contains 4 white, 5 red and 6 black balls. Three are drawn at random. Find the probability that (i) no ball drawn is black, (ii) exactly 2 are black (iii) all are of the same colour.

• A. $\frac{12}{65},\frac{27}{91},\frac{6}{91}$
• B. $\frac{12}{65},\frac{27}{91},\frac{24}{455}$
• C. $\frac{12}{65},\frac{27}{91},\frac{14}{455}$
• D. $\frac{12}{65},\frac{27}{91},\frac{34}{455}$

Answer 1

Answer: (D)

$\frac{12}{65},\frac{27}{91},\frac{34}{455}$

Total number of balls $=15$
Total number of ways of selecting $3$ balls out of $15$ balls is ${}^{15}{C}_3$.
Number of black balls $=6$
Number of non-black balls $=9$ (4 white, 5 red)
Probability of no ball drawn is black $=\frac{{}^9{C}_3}{{}^{15}{C}_3}=\frac{12}{65}$
Probability of exactly 2 black $=\frac{{}^6{C}_2{\times }^9{C}_1}{{}^{15}{C}_3}=\frac{27}{91}$
Probability of all of same colour i.e. $RRRorWWWorBBB$
$=\frac{{}^4{C}_3{+}^5{C}_3{+}^6{C}_3}{{}^{15}{C}_3}$
$=\frac{4+10+20}{\frac{\left(\mathrm{15.14.13}\right)}{(3!)}}=\frac{34}{455}$