Example 15 - Find probability that when a hand of 7 cards is - Using combination

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  1. Chapter 16 Class 11 Probability
  2. Serial order wise

Transcript

Example 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 7 cards are to be chosen from 52 cards Total number of combinations (hands) possible = 52C7 = 52! 7! 52 7 ! = 52! 7!45! Let A be the event that all kings are selected There are only 4 kings in a pack of 52 cards Hence if 7 cards are chosen 4 kings to be chosen out of 4 and 3 others to be chosen out of remaining 48 Hence total number of combinations n(A) = 4C4 48C3 = 4! 4!0! 48! 3! 48 3 ! = 1 48 3! 45! = 48! 3! 45! Hence P (A) = ( ) ( ) = 48! 3! 45! 52! 7! 45! = 48! 7! 3! 52! = Example 15 Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (ii) 3 Kings Let B be that event that 3 king are selected There are only 4 king in a pack of 52 cards Hence if 7 cards are chosen, 3 king to be chosen out of 4 and 4 other to be chosen out of remaining 48 Hence, number of combination n(B) = 4C3 48C4 = 4! 3! 4 3 ! 48! 4! 48 4 ! = 4 48! 4!44! Hence, P(B) = ( ) ( ) = 4 48! 4!44! 52! 7!45! = 4 48! 4!44! 7!45! 52! = 4 48! 7! 45! 4! 44! 52! = 4 48! 7! 45! 4! 44! 52 51 50 49 48! = 4 7! 45 4! 52 51 50 49 = 4 7 6 5 4! 45 52 51 50 49 4! = Example 15 Find the probability that when a hand of 7 cards is drawn from a well shuffled deck of 52 cards, it contains (iii) at least 3 Kings. Atleast 3 kings are selected means either 3 kings are selected or 4 kings are selected So, P(at least 3 king) = P(3 King) + ( 4 King) We know that, P(3 Kings) = 9 1547 P(4 Kings) = 1 7735 P(at least 3 king) = 9 1547 + 1 7735 =

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