Question

An urn contains $10$ balls coloured either black or red When selecting two balls from the urn at random, the probability that a ball of each color is selected is $8/15$. Assuming that the urn contains more black balls then red balls, the probability that at least one black ball is selected, when selecting two balls, is

• A. $\frac{18}{45}$
• B. $\frac{30}{45}$
• C. $\frac{39}{45}$
• D. $\frac{41}{45}$

Answer 1

Answer: (C)

$\frac{39}{45}$

Number of black balls $=x$

Number of red balls $=y$

$i)=x+y=10$ (Total)

$ii)P$ (selecting exactly $1$ black & one red)

${=}^x{C}_1{\times }^y{C}_1{/}^{10}{C}_2=8/15$

by equation : $x^2-10x+24=0;x=4$ or $6$

since $x>y$ it is $6$

$iii)P$ (selecting at least one black)

${=}^x{C}_1{\times }^y{C}_1{+}^x{C}_2{/}^{10}{C}_2$

Above equation reduced to $\Rightarrow 2xy+x(x-1)/90$

putting $x=6$, results is $39/45$