# Ex 13.3, 12 - Chapter 13 Class 12 Probability (Term 2)

Last updated at Feb. 15, 2020 by Teachoo

Last updated at Feb. 15, 2020 by Teachoo

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(๐ธ_1 ).๐(๐ด|๐ธ_1))/(๐(๐ธ_1 ).๐(๐ด|๐ธ_1)+๐(๐ธ_2 ).๐(๐ด|๐ธ_2) ) "P(E1)" = Probability that lost card is diamond = 13/52 = ๐/๐ P(A|E1) = Probability of getting 2 diamond cards if lost card is diamond = โ(๐๐๐๐๐๐ก๐๐๐ 2 ๐๐๐๐๐๐๐ ๐๐๐๐๐ ๐๐๐๐ @(13โ1= 12)๐๐๐๐๐๐๐ ๐๐๐๐๐ )/(๐๐๐๐๐๐ก๐๐๐ ๐๐๐ฆ 2 ๐๐๐๐๐ ๐๐๐๐ 51 ๐๐๐๐๐ ) = ๐๐๐ช๐/๐๐๐ช๐ = ((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 = ๐/๐ P(A|E2) = Probability of getting 2 diamond cards if lost card is not a diamond = โ(๐๐๐๐๐๐ก๐๐๐ 2 ๐๐๐๐๐๐๐ ๐๐๐๐๐ @๐๐๐๐ 13 ๐๐๐๐๐๐๐ ๐๐๐๐๐ )/(๐๐๐๐๐๐ก๐๐๐ ๐๐๐ฆ 2 ๐๐๐๐๐ ๐๐๐๐ 51 ๐๐๐๐๐ ) = ๐๐๐ช๐/๐๐๐ช๐ = ((13 ร 12)/2!)/((51 ร 50)/2!) = (13 ร 12)/(51 ร 50) Putting value in the formula, P(E1|A) = (๐(๐ธ_1 ).๐(๐ด|๐ธ_1))/(๐(๐ธ_1 ).๐(๐ด|๐ธ_1)+๐(๐ธ_2 ).๐(๐ด|๐ธ_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) = ๐๐/๐๐ Therefore, required probability is 11/50

Chapter 13 Class 12 Probability (Term 2)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.