Please provide specific answer 6. (+) 4 cards are randomly dealt to each of 13 p

Question

Please provide specific answer

6. (+) 4 cards are randomly dealt to each of 13 players. (a) Describe the sample space and write down its size. (b) What is the probability of A- feach player has one card from each suit? (c) What is the probability of {each player has all four cards of the same value? (d) * Compute PB(A) and PA(B). Are A and B independent? (e) What is the probabiility that players one, two, and three have been dealt only aces, kings, or queens. (f) (**) What is the probability that one player has one card from each suit but that nobody As usual, explain. else has cards from more than one suit?

Explanation / Answer

A. Total Numbe r of cards in a pack= 52

Number of ways in which 53 Cards is distributed among 13 Players = 52 C13

After 13 cards have been distributed remaining is 39 cards so these 39 cards is distributed among 13 players in 39C13

Likewise 26C13, 13C13

So there are 4 groups and these groups being interchangable

So total Sample set= (52P13)

B> Each Player gets one card from each suit

So there are 4 Suits and each suits comprises of 13 cards each

Out 4 suits 1 can be chosen in 4C1 ways and the 13 cards crd be distributed in 13P13 ways

So P(A)= [4C1*13P13*3C1*13P13*2C1*13P13+1C1*13P13]/ [52P13]= 4!/52P13

C> In pack of 52 Cards there are cards of 13 Values

So out 13 Values one can be chosen is 13C1 and distributed to 1 Player in 1way

So possible ways= [13C1*4P4*12C1*4P4*11C1*4P4.....1c1*4P4]
So probability = [13C1*$p4*12c1*4p4*..........1c1*4p4]/ [ 52p13]= 13!/52P13

P stands for permutations

E> In a pack there 4 caes, 4 kings and 4 queens

So keeping aside these 12 cards remaining are 40 Cards

So 40 cards is distributed among 10 players as 1,2 & 3 players are kept aside in 40P10 ways

Ace given player 1 way, Player 2 Player 3 recaive only aces king and queens

P(e)= 1- [ 40p10/(52p13)

F >Same as that B

D> P(A/B)= {P(B/A)*P(A)}/{P(B)}; P(B/A)={P(A/B)*P(B)/PA)} (1)

So we know that each suit has 13 cards/Values and there 4 suits

So probability of one p[layer having card from each suit and all the cards of the same value in 1 ways

So P(b/A)= P(A/B)=1

So replacing int eqn(1) we get that both the events are independent

End of answer!

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