Question

In an international film festival, a penal of 11 judges is formed to judge the best film. At last two films FA and FB were considered to be the best where the opinion of judges got divided. Six judges where in favor of FA whereas five in favor of FB. A random sample of five judges was drawn from the panel. Find the probability that out of five judges, three are in favor of film FA.

Answer 1

Answer:

The answer is "0.4329 ".

Step-by-step explanation:

P( three in favor of FA)

Select 3 out of 6 FA supporters then select 2 out of 5 FB supportive judges  

$=\frac{^{6}_{C_{3}}\times ^{5}_{C_{2}}}{^{11}_{C_{5}}}\\\\=\frac{\frac{6!}{3!(6-3)!}\times \frac{5!}{2!(5-2)!}}{\frac{11!}{5!(11-5)!}}\\\\=\frac{\frac{6!}{3! \times 3!}\times \frac{5!}{2! \times 3!}}{\frac{11!}{5! \times 6!}}\\\\=\frac{\frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1\times 3!}\times \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!}}{\frac{11 \times10 \times 9 \times 8 \times 7 \times 6! }{5 \times 4 \times 3 \times 2 \times 1 \times 6!}}$

$=\frac{ (5 \times 4) \times(5 \times 2)}{(11 \times 3 \times 2 \times 7 )}\\\\=\frac{ 20 \times 10 }{(11 \times 42)}\\\\=\frac{ 200 }{462}\\\\=\frac{100 }{231}\\\\=0.4329$