Question

7. (S 2.3) A hand of 13 cards is to be dealt at random and without replacement from an ordinary deck of playing cards. Find the conditional probability that there are four kings in the hand given that the hand contained at least one king. 8. (S 2.3) In a certain factory, machines I, II, and III are all producing springs of the same length. Machines 1, 11, and 111 produce 1%, 4%, and 2% defective springs, respectively. Of the total production of springs in the factory, Machines 1 produces 30%, Machines 11 produces 25%, and Machines 111 produces 45%. (a) If one spring is selected at random from the total springs produced in a given day determine the probability that it is defective. (Hint: Total Law of Probability) (b) Given that the selected spring is defective, find the conditional probability that it was produced by Machine II. (Hint: Bayes' Theorem)

Answer 1

7)P(4 king in 13 cards)=P(4 king from 4 kings and 9 cards from rest of 48)

= $\small \binom{4}{4}*\binom{48}{9}/\binom{52}{13}$ =1*1677106640/635013559600

P(at least one king)=1-P(no king)=1-P(all 13 cards from 48 non king cards)=1- $\small \binom{48}{13}/\binom{52}{13}$

=1-192928249296/635013559600=442085310304/635013559600

hence P(4 king given at least one king)=

=(1*1677106640/635013559600)/(442085310304/635013559600)=0.003794

8)

a)P(defective)=P(machine I and defective)+P(machine II and defective)+P(machine III and defective)

=0.3*0.01+0.25*0.04+0.45*0.02=0.022

b)P(machine II given defective)=P(machine II and defective)/P(defective)

=0.25*0.04/0.022=0.4545