Many popular games are based on a standard deck of 52 playing cards. A face card

Question

Many popular games are based on a standard deck of 52 playing cards. A face card haw a value and a suit. There are 13 face cards values: {A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K} and each lace card comes from one of 4 suits {Hearts, Diamonds, Clubs, Spades). After shuffling two standard 52 card decks separately, one card is drawn from the first 52 deck is placed in the second deck. A card is then drawn randomly from the second deck. (a) What is the probability that the second card is an Ace? (b) Given that the second card drawn is an Ace, what is the conditional probability that an ace was transferred from the first deck of cards to the second? Let P(A) = 0.4 and P(A B) = 0.6. (a) If A and B are mutually exclusive, then what is P(B) =? (b) If A and B are independent, then what is P(B)? If A, B and C are events such that P(A) - 3/3, P(B) = 1/4 and P(C) = 3/5. Find the P(A B C) under the following assumptions: (a) If A, B and C are mutually exclusive. (b) If A, B and C are independent.

Explanation / Answer

7(a) P( Drawing a Ace Card from the Second Deck of cards) = P( Drawing a Ace card when Non Ace card was transfered from Deck 1 to deck 2)+P( Drawing a Ace card when an Ace card was transfered from deck 1 to deck 2)

=4C1/53C1 + 5C1/53C1 =9/53

(b) Let A be the event of Drawing a Ace card from Deck !

   Let B be the event of drawing a Ace card from II deck

To find P(A/B)= [P(B/A)P(B) ]/[P(B/A)P(A)+P(B/A')P(A')] --------------(Bayes theorem)

       here P(B/A)=5/53,P(A)=4/52,P(B/A')=4/53,P(A')=48/52

so we get P(A/B)=5/53

(c) P(A)=0.4

P(AUB)=0.6

8 (a) If A and B are mutually exclusive then P(A n B)=0

so P(AUB)=P(A)+P(B)

=>0.6=0.4+P(B)

=>P(B)=0.20

(b) If A and B are Independent, then P(A n B)=P(A).P(B)

so P(AUB)=P(A)+P(B)-P(A).P(B)

=>0.6=0.4+P(B)-0.4P(B)

=>0.20=P(B)(0.60)

=>P(B)=0.20/0.60=1/3

9(a)given P(A)=1/3 ,P(B)=1/4,P(C)=1/5,

if A ,B ,C are mutually exclusive, then P(AUBUC)=P(A)+P(B)+P(C)=47/60

(b) IF A,B,C are Independent, then P(AnB)=P(A)P(B),P(AnC)=P(A)P(C),P(Bnc)=P(B)P(C),P(AnBnC)=P(A)P(B)P(C)

so P(AUBUC)=P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC)=3/5

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