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

Marks: 6M

Year: May 2015

A dice can show number 3 or 5 or it will not.

So it is problem of binomial distribution.

Let X denote number of dice showing three or five.

Let P = probability to show number 3 or 5 = 2/6 = 1/3

$\mathrm{\therefore }$ q = 1-p = 1 - 1/3 = 2/3

N = 7 and N = 729

$$\mathrm{\therefore } P(X= x) = n_{C_x}\ p^x\ q^{n-x}$$ $$ = n_{C_x}\ {(\frac{1}{3})}^x\ {(\frac{2}{3})}^{n-x}$$

Probability to show at least 4 dice three or five : $$P(X \mathrm{\ge} 4)$$ $$= \sum^7_{x=4}{7_{C_x}\ {(\frac{1}{3})}^x\ {(\frac{2}{3})}^{7-x}}$$ $$= 7_{C_4}\ {(\frac{1}{3})}^4\ {(\frac{2}{3})}^3+\ 7_{C_5}\ {(\frac{1}{3})}^5\ {(\frac{2}{3})}^2 + 7_6\ {(\frac{1}{3})}^6\ {(\frac{2}{3})}^1+\ 7_{C_7}\ {(\frac{1}{3})}^7\ {(\frac{2}{3})}^0$$ $$= 35 * 1/81 * 8/27 + 21 * 1/243 * 4/9 + 7 * 1/729 *2/3 + 1* 1/2187$$

$$= 379/2187$$

$\mathrm{\therefore }$ Number of times at least four dice show 3 or 5: $$N X P(X \mathrm{\ge} 4)$$

$$= 729 *(379/2187)$$

$$= 126$$

Thus for 126 times we can expect at least four dice to show three or five.