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  1. Chapter 13 Class 12 Probability
  2. Serial order wise

Transcript

Ex 13.3, 12 A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.Let, E1 : Event that lost card is diamond E2 : Event that lost card is not a diamond A : Event that two cards drawn are diamond We need to find out that probability that the lost card being a diamond if two cards drawn are found to be both diamond. i.e. P(E1|A) P(E1|A) = (๐‘ƒ(๐ธ_1 ).๐‘ƒ(๐ด|๐ธ_1))/(๐‘ƒ(๐ธ_1 ).๐‘ƒ(๐ด|๐ธ_1)+๐‘ƒ(๐ธ_2 ).๐‘ƒ(๐ด|๐ธ_2) ) "P(E1)" = Probability that lost card is diamond = 13/52 = ๐Ÿ/๐Ÿ’ P(A|E1) = Probability of getting 2 diamond cards if lost card is diamond = โ–ˆ(๐‘†๐‘’๐‘™๐‘’๐‘๐‘ก๐‘–๐‘›๐‘” 2 ๐‘‘๐‘–๐‘Ž๐‘š๐‘œ๐‘›๐‘‘ ๐‘๐‘Ž๐‘Ÿ๐‘‘๐‘  ๐‘“๐‘Ÿ๐‘œ๐‘š @(13โˆ’1= 12)๐‘‘๐‘–๐‘Ž๐‘š๐‘œ๐‘›๐‘‘ ๐‘๐‘Ž๐‘Ÿ๐‘‘๐‘ )/(๐‘†๐‘’๐‘™๐‘’๐‘๐‘ก๐‘–๐‘›๐‘” ๐‘Ž๐‘›๐‘ฆ 2 ๐‘๐‘Ž๐‘Ÿ๐‘‘๐‘  ๐‘“๐‘Ÿ๐‘œ๐‘š 51 ๐‘๐‘Ž๐‘Ÿ๐‘‘๐‘ ) = ๐Ÿ๐Ÿ๐‘ช๐Ÿ/๐Ÿ“๐Ÿ๐‘ช๐Ÿ = ((12 ร— 11)/2!)/((51 ร— 50)/2!) = (12 ร— 11)/(51 ร— 50) "P(E2)" = Probability that lost card is not a diamond = 1 โ€“ P(E1) = 1 โ€“ 1/4 = ๐Ÿ‘/๐Ÿ’ P(A|E2) = Probability of getting 2 diamond cards if lost card is not a diamond = โ–ˆ(๐‘†๐‘’๐‘™๐‘’๐‘๐‘ก๐‘–๐‘›๐‘” 2 ๐‘‘๐‘–๐‘Ž๐‘š๐‘œ๐‘›๐‘‘ ๐‘๐‘Ž๐‘Ÿ๐‘‘๐‘  @๐‘“๐‘Ÿ๐‘œ๐‘š 13 ๐‘‘๐‘–๐‘Ž๐‘š๐‘œ๐‘›๐‘‘ ๐‘๐‘Ž๐‘Ÿ๐‘‘๐‘ )/(๐‘†๐‘’๐‘™๐‘’๐‘๐‘ก๐‘–๐‘›๐‘” ๐‘Ž๐‘›๐‘ฆ 2 ๐‘๐‘Ž๐‘Ÿ๐‘‘๐‘  ๐‘“๐‘Ÿ๐‘œ๐‘š 51 ๐‘๐‘Ž๐‘Ÿ๐‘‘๐‘ ) = ๐Ÿ๐Ÿ‘๐‘ช๐Ÿ/๐Ÿ“๐Ÿ๐‘ช๐Ÿ = ((13 ร— 12)/2!)/((51 ร— 50)/2!) = (13 ร— 12)/(51 ร— 50) Putting value in the formula, P(E1|A) = (๐‘ƒ(๐ธ_1 ).๐‘ƒ(๐ด|๐ธ_1))/(๐‘ƒ(๐ธ_1 ).๐‘ƒ(๐ด|๐ธ_1)+๐‘ƒ(๐ธ_2 ).๐‘ƒ(๐ด|๐ธ_2) ) = (1/4 ร— (12 ร— 11)/(51 ร— 50))/( 1/4 ร— (12 ร— 11)/(51 ร— 50) + 3/4 ร— (13 ร— 12)/(51 ร— 50) ) = (1/4 ร— (12 )/(51 ร— 50) (11))/( 1/4 ร— 12/(51 ร— 50) (11+ 3 ร— 13) ) = 11/(11 + 39) = ๐Ÿ๐Ÿ/๐Ÿ“๐ŸŽ Therefore, required probability is 11/50

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Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.