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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
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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) = ﷐𝑛(𝐴)﷮𝑛(𝑆)﷯ = ﷐48!﷮3! 45!﷯ ÷ ﷐52!﷮7! 45!﷯ = ﷐48! × 7!﷮3! × 52!﷯ = ﷐𝟏﷮𝟕𝟕𝟑𝟓﷯ 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) = ﷐𝑛(𝐵)﷮𝑛(𝑆)﷯ = ﷐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!﷯ = ﷐𝟗﷮𝟏𝟓𝟒𝟕﷯ 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﷯ = ﷐𝟒𝟔﷮𝟕𝟕𝟑𝟓﷯

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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