Question

Suppose that a central university has to form a committee of $5$ members from list of $20$ candidates out of whom $12$ are teacher and $8$ are students .If the member of the committee are selected random,what is the probability that the majority of the committee members are students ?

Answer 1

Given Total Number of Candidates $=20$

Teacher $=12$

Student $=8$

The cases when majority will be student are

$\to$ student $=5$, teacher $=0$

$\to$ student $=4$, teacher $=1$

$\to$ student $=3$, teacher $=2$

Total probability is $\frac{{}^8{C_5}^{12}{C}_0{+}^8{C_4}^{12}{C}_1{+}^8{C_3}^{12}{C}_2}{{}^8{C_5}^{12}{C}_0{+}^8{C_4}^{12}{C}_1{+}^8{C_3}^{12}{C}_2{+}^8{C_2}^{12}{C}_3{+}^8{C_1}^{12}{C}_{11}{+}^{12}{C}_5}$

$\Rightarrow \frac{\frac{8\times 7\times 6}{3\times 2}\times 1+\frac{8\times 7\times 6\times 5}{4\times 3\times 2}\times 12+\frac{8\times 7\times 6}{3\times 2}\times \frac{12\times 11}{2}}{\frac{8\times 7\times 6}{3\times 2}\times 1+\frac{8\times 7\times 6\times 5}{4\times 3\times 2}\times 12+\frac{8\times 7\times 6}{3\times 2}\times \frac{12\times 11}{2}+\frac{8\times 7}{2}\times \frac{12\times 10\times 11}{3\times 2}+\frac{8\times 12\times 11\times 10\times 89}{4\times 3\times 2}+\frac{12\times 11\times 10\times 9\times 8}{5\times 4\times 3\times 2}}$

$\Rightarrow \frac{56+70\times 12+56\times 66}{56+70\times 12+56\times 66+28\times 220+3960+792}$

$\Rightarrow \frac{56+840+3696}{56+840+3696+6160+3960+792}\Rightarrow \frac{4592}{15,504}=0.296$