Mathematics.....................................

Question

In a certain city only 50% of students are capable of doing college work actually go to college. Assuming that this claim is true, find the probability that among 18 such a capable students (i) exactly 10 will go to college (ii) at least 2 will go to college (iii) at most 17 will go to college?

Answer 1

\(p = 50\% = \frac{50}{100} = \frac 12\)

\(q = 1 - p = 1 - \frac 12 = \frac 12\)

\(n = 18\)

(i) \(P(x = 10) = \,^{18} C_{10} \,p^{10} q^8\) 

\(=\, ^{18}C_{10} \left(\frac 12\right)^{10}\left(\frac 12\right)^{8}\)

\(=\left(\frac 12\right)^{18} \,^{18}C_{10}\)

(ii) \(P(x \ge 2) = 1 - P(x = 0) - P(x = 1)\) 

\(= 1 -\, ^{18}C_0\,p^0 q^{18} -\, ^{18}C_1 \,p^1q^{17}\)

\(= 1 - \left(\frac 12\right)^0 \left(\frac 12\right)^{18} - 18\left(\frac 12\right) \left(\frac 12\right)^{17}\)

\(= 1 -\left(\frac 12\right)^{18} - 18\left(\frac 12\right)^{18}\)

\(= 1- 19\left(\frac12\right)^{18}\)

(iii) \(P(X \le 17) = 1 - P(X = 18)\)

\(= 1-\,^{18}C_{18} \,p^{18}q^0\)

\(= 1- \,^{18}C_{18}\left(\frac 12\right)^{18} \left(\frac 12\right)^0\)

\(=1-\left(\frac 12\right)^{18}\)