Question

There are 8 brown balls, 4 oragne balls and 5 black balls in a bag. Five balls are chosen at random. What is the probability of there being 2 brown balls, 1 orange ball and 2 black balls?

• A. $\frac{191}{1547}$
• B. $\frac{180}{1547}$
• C. $\frac{280}{1547}$
• D. $\frac{180}{1547}$
• E. None of these

Answer 1

The correct option is C $\frac{280}{1547}$

Toal possible outcomes ${=}^{17}{C}_5$
$=\frac{17\times 16\times 15\times 14\times 13}{1\times 2\times 3\times 4\times 5}=6188$
Total favourable outcomes ${=}^8{C}_2{\times }^4{C}_1{\times }^5{C}_2$
$=\frac{8\times 7}{1\times 2}\times 4\times \frac{5\times 4}{1\times 2}=28\times 4\times 10=1120$
Required probability $=\frac{1120}{6188}=\frac{280}{1547}$