Question:
15 Find the probability that when a hand of 7 cards is drawn from a well
shuffled deck of 52 cards, it contains (i) all Kings (ii) 3 Kings (iii) atleast 3 Kings.
Answer:
Total number of cards = 52
Number of king card = 4
Now, 7 cards are drawn from 52 cards.
1. P (all cards are king) = { 4C4 * 48C3 }/ 52C7
= {(48*47*46)/(3*2*1)}/{(52*51*50*49*48*47*46)/(7*6*5*4*3*2*1)}
= {(48*47*46)*(7*6*5*4*3*2*1)}/{(3*2*1)*{(52*51*50*49*48*47*46)}
= (7*6*5*4)/(52*51*50*49)
= (6*5*4)/(52*51*50*7)
= (6*5)/(7*13*51*50)
= 6/(7*52*51*10)
= 2/(7*13*17*10)
= 1/(7*13*17*5)
= 1/7735
2.P (3 cards are king) = { 4C3 * 48C4 }/ 52C7
= {4*(48*47*46*45)/(4*3*2*1)}/{(52*51*50*49*48*47*46)/(7*6*5*4*3*2*1)}
= {4*(48*47*46*45)*(7*6*5*4*3*2*1)}/{(4*3*2*1)*{(52*51*50*49*48*47*46)}
= (7*6*5*4*45)/(52*51*50*49)
= (6*5*4*45)/(52*51*50*7)
= (6*4*45)/(7*52*51*10)
= (6*45)/(7*13*51*10)
= (6*3)/(7*13*17*2)
= (3*3)/(7*13*17)
= 9/1547
3. P (all cards are king) = P (4 cards are king) + P (3 cards are king)
= { 4C4 * 48C3 }/ 52C7 + { 4C3 * 48C4 }/ 52C7
= (9/1547) + (1/7735)
= 46/7735
Not what you are looking for? Go ahead and submit the question, we will get back to you.
