Question

A bag contains some white and some black balls, all of which are distinguishable from each other, all combinations of balls being equally likely. The total number of balls in the bag is $10$. If three balls are drawn at random and all of them are found to be black, the probability that the bag contains $1$ white and $9$ black balls is :

• A. $\frac{14}{55}$
• B. $\frac{12}{55}$
• C. $\frac{8}{55}$
• D. $\frac{2}{11}$

Answer 1

The correct option is A $\frac{14}{55}$

Possible cases are:
(i) $3B,7W$; (ii) $4B,6W$; (iii) $5B,5W$; (iv) $6B,4W$;
(v) $7B,3W$; (vi) $8B,2W$; (vii) $9B,1W$; (viii) $10B,0W$
Probability of each case $=\frac{1}{8}$
Probability of getting $3B$ in the above cases:
(i) $\frac{{}^3{C}_3}{{}^{10}{C}_3}$; (ii) $\frac{{}^4{C}_3}{{}^{10}{C}_3}$; (iii) $\frac{{}^5{C}_3}{{}^{10}{C}_3}$; (iv) $\frac{{}^6{C}_3}{{}^{10}{C}_3}$
(v) $\frac{{}^7{C}_3}{{}^{10}{C}_3}$; (vi) $\frac{{}^8{C}_3}{{}^{10}{C}_3}$; (vii) $\frac{{}^9{C}_3}{{}^{10}{C}_3}$; (viii) $\frac{{}^{10}{C}_3}{{}^{10}{C}_3}$
Therefore, required probability is
$\frac{\frac{{}^9{C}_3}{{}^{10}{C}_3}\times \frac{1}{8}}{(\frac{{}^3{C}_3}{{}^{10}{C}_3}+\frac{{}^4{C}_3}{{}^{10}{C}_3}+\frac{{}^5{C}_3}{{}^{10}{C}_3}+\frac{{}^6{C}_3}{{}^{10}{C}_3}+\frac{{}^7{C}_3}{{}^{10}{C}_3}+\frac{{}^8{C}_3}{{}^{10}{C}_3}+\frac{{}^9{C}_3}{{}^{10}{C}_3}+\frac{{}^{10}{C}_3}{{}^{10}{C}_3})\frac{1}{8}}=\frac{84}{330}=\frac{14}{55}$
Hence, A is correct.