Question

In a particular university 40% of the students are having newspaper reading habit. Nine university students are selected to find their views on reading habit. Find the probability that atleast two-third have newspaper reading habit

Answer 1

Let p to the probability of having newspaper reading habit

p = ${\displaystyle \frac{40}{100}=\frac{2}{5}}$

q = 1 – p

= ${\displaystyle 1\frac{2}{5}}$

= ${\displaystyle \frac{5-2}{5}}$

= ${\displaystyle \frac{3}{5}}$ and n = 9

In the binomial distribution p(x = 4) = nc_x p^xq^n-r 

The binomial distribution P(x) = ${\displaystyle 9{\text{C}}_x{\left(\frac{2}{5}\right)}^x{\left(\frac{3}{5}\right)}^{9-x}}$

P(at least two third have newspaper reading habit)

${\displaystyle \text{P}(x\ge 9\times \frac{2}{3})}$

= ${\displaystyle \text{P}(x\ge 6)}$

 = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

= ${\displaystyle 9{\text{c}}_6{\left(\frac{2}{5}\right)}^6{\left(\frac{3}{5}\right)}^{9-6}+9{\text{c}}_7{\left(\frac{2}{7}\right)}^7{\left(\frac{3}{5}\right)}^{9-7}+9{\text{c}}_8{\left(\frac{2}{5}\right)}^8{\left(\frac{3}{5}\right)}^{9-8}+9{\text{c}}_9{\left(\frac{2}{5}\right)}^9{\left(\frac{3}{5}\right)}^{9-9}}$

= ${\displaystyle 9{\text{c}}_3{\left(\frac{2}{5}\right)}^6{\left(\frac{3}{5}\right)}^3+9{\text{c}}_2{\left(\frac{2}{5}\right)}^7{\left(\frac{3}{5}\right)}^2+9{\text{c}}_1{\left(\frac{2}{5}\right)}^8{\left(\frac{3}{5}\right)}^1+1{\left(\frac{2}{5}\right)}^9{\left(\frac{3}{5}\right)}^0}$

= ${\displaystyle \frac{9\times 8\times 7}{1\times 2\times 3}\times \left[\frac{{\left(2\right)}^6\times {\left(3\right)}^3}{{\left(5\right)}^9}\right]+\frac{9\times 8}{1\times 2}\left[\frac{{\left(2\right)}^7\times {\left(3\right)}^2}{{\left(5\right)}^9}\right]+9\times \left[\frac{{\left(2\right)}^8\times 3}{{\left(5\right)}^9}\right]+\frac{{\left(2\right)}^9}{{\left(5\right)}^9}}$

= ${\displaystyle 84\times \left(\frac{64\times 27}{{\left(5\right)}^9}\right)+36\left[\frac{128\times 9}{{\left(5\right)}^9}\right]+9\times \left[\frac{256\times 3}{{\left(5\right)}^9}\right]+\left[\frac{512}{{\left(5\right)}^9}\right]}$

= ${\displaystyle \frac{145152}{{\left(5\right)}^9}+\frac{41472}{{\left(5\right)}^9}+\frac{6912}{{\left(5\right)}^9}+\frac{512}{{\left(5\right)}^9}}$

= ${\displaystyle \frac{145152+41472+6912+512}{{\left(5\right)}^9}}$

= ${\displaystyle \frac{194048}{195312}}$

= 0.09935