Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: May 2014

For binomial distribution , n=5

$$P(X = x) = 5_{C_2\ ({0.3)}^x\ ({0.7)}^{5-x}}$$

Given : $$\frac{P\left(X=3\right)}{P\left(X=2\right)}=\frac{1}{4}$$ $$\boldsymbol{\mathrm{\therefore }} 4 P(X=3) = P(X=2)$$ $$4 * 5_{C_3}\ p^3\ q^{5-3}=5_{C_2}\ p^2\ q^{5-2}\ $$ $$\mathrm{\therefore } 4* 10 p^3\ q^2 = 10p^2\ q^3$$ $$4p = q$$

$$\mathrm{\therefore } 4p = 1-p$$

$$\mathrm{\therefore } 5p = 1$$

$$\mathrm{\therefore } p =1/5$$

$$\mathrm{\therefore } q = 1 - 1/5 = 4/5$$

Part 2 :

When P= 1/5 , q = 4/5 , n=6

P(4 successes) $$P(X = 4)$$ $$ = 6_{C_4\ }({1/5)}^4\ {(\frac{4}{5})}^{6-4}$$ $$ = 15 * 1/625 * 16/25 = 48/3125$$