A has 3 shares in a lottery in which there are 3 prizs and 8 blanks B has 2 - Maths - Probability

Question

A has 3 shares in a lottery in which there are 3 prizs and 8
blanks B has 2 shares in a lottery in which the are 4 prizes and 6
blanks compar their chances of success

Ajanta Trivedi · Accepted Answer

since A has three shares in which 3 prizes and 8 blanks are there.

A can win in three ways, either he get 1 prize and 2 blank or 2 prizes and 1 blank or all the three prizes.

All the three cases where 'A' can win is independent and mutually exclusive. hence probability that A can win prize is given as;

$P (A) = \frac{{}^{3}{C_{1}} \cdot {}^{8}{C_{2}}}{{}^{11}{C_{3}}} + \frac{{}^{3}{C_{2}} \cdot {}^{8}{C_{1}}}{{}^{11}{C_{3}}} + \frac{{}^{3}{C_{3}}}{{}^{11}{C_{3}}} {}^{11}{C_{3}} = \frac{11 * 10 * 9}{3 * 2} = 11 * 5 * 3 = 165 {}^{8}{C_{2}} = \frac{8 * 7}{2} = 28 P (A) = \frac{3 * 28}{165} + \frac{3 * 8}{165} + \frac{1}{165} = \frac{84 + 24 + 1}{165} = \frac{109}{165}$

B has 2 shares in a lottery in which 4prizes and 6 blanks are there.

B can win in two ways , either he gets 1 prize and 1 blank or 2 prizes.

$P (B) = \frac{{}^{4}{C_{1}} \cdot {}^{6}{C_{1}}}{{}^{10}{C_{2}}} + \frac{{}^{4}{C_{2}}}{{}^{10}{C_{2}}} {}^{4}{C_{2}} = \frac{4 * 3}{2} = 6 {}^{10}{C_{2}} = \frac{10 * 9}{2} = 45 P (B) = \frac{4 * 6}{45} + \frac{6}{45} = \frac{24 + 6}{45} = \frac{30}{45} = \frac{2}{3}$

$P (A) : P (B) = \frac{109}{165} : \frac{2}{3} = \frac{109}{165} * \frac{3}{2} = \frac{109}{55 * 2} = \frac{109}{110}$

A has 3 shares in a lottery in which there are 3 prizs and 8 blanks. B has 2 shares in a lottery in which the are 4 prizes and 6 blanks. compar their chances of success.