Answer to Question #187013 in Statistics and Probability for Chairmane

Question

 From a box of a dozen rocket-propelled grenade (RPG), 4 are selected at random and fired. If the box contains 3 defective RPG’s, what is the probability that (a) all 4 will fire? (b) at most 2 will not fire? 

Answer 1

p - RPG fire, q - RPG doesn't fire

a)

$P(p=4)=\frac{9}{12}\cdot\frac{8}{11}\cdot\frac{7}{10}\cdot\frac{6}{9}=\frac{7\cdot2}{11\cdot5}=\frac{14}{55}$

b)

$P(q\le2)=P(q=0)+P(q=1)+P(q=2)=$

$=\frac{3}{12}\cdot\frac{9}{11}\cdot\frac{8}{10}\cdot\frac{7}{9}+\frac{3}{12}\cdot\frac{2}{11}\cdot\frac{9}{10}\cdot\frac{8}{9}=\frac{7}{55}+\frac{2}{55}=\frac{9}{55}$