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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 22 Coloured balls are distributed in four boxes as shown in the following table: A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the boxs III?Let A : Event that a black ball is selected E1 : Event that the ball is selected from box I E2 : Event that the ball is selected from box II E3 : Event that the ball is selected from box III E4 : Event that the ball is selected from box IV We need to find out the Probability of the ball drawn is from box III if it is black. i.e. P(E3|A) P(E3|A)= (๐‘ƒ(๐ธ3).๐‘ƒ(๐ด|๐ธ3))/(๐‘ƒ(๐ธ1)๐‘ƒ(๐ด|๐ธ1)+๐‘ƒ(๐ธ2)๐‘ƒ(๐ด|๐ธ2)+๐‘ƒ(๐ธ3)๐‘ƒ(๐ด|๐ธ3)+๐‘ƒ(๐ธ4)๐‘ƒ(๐ด|๐ธ4)) P(E1) = Probability that ball drawn is from box I = ๐Ÿ/๐Ÿ’ P(A|E1) = Probability of that black ball is selected from Box I = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘Ž๐‘™๐‘™๐‘ )/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘Ž๐‘™๐‘™๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘๐‘œ๐‘ฅ) = ๐Ÿ‘/๐Ÿ๐Ÿ– P(E2) = Probability that ball drawn is from box II = ๐Ÿ/๐Ÿ’ P(A|E2) = Probability of that black ball is selected from Box II = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘Ž๐‘™๐‘™๐‘ )/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘Ž๐‘™๐‘™๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘๐‘œ๐‘ฅ) = 2/8 = 1/4 P(E3) = Probability that ball drawn is from box III = ๐Ÿ/๐Ÿ’ P(A|E3) = Probability of that black ball is selected from Box II = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘Ž๐‘™๐‘™๐‘ )/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘Ž๐‘™๐‘™๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘๐‘œ๐‘ฅ) = ๐Ÿ/๐Ÿ• P(E4) = Probability that ball drawn is from box IV = ๐Ÿ/๐Ÿ’ P(A|E4) = Probability of that black ball is selected from Box IV = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘Ž๐‘™๐‘™๐‘ )/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘Ž๐‘™๐‘™๐‘  ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘๐‘œ๐‘ฅ) = ๐Ÿ’/๐Ÿ๐Ÿ‘ Putting values in Equation "P(E3|A)"=(๐‘ƒ(๐ธ3).๐‘ƒ(๐ด|๐ธ3))/(๐‘ƒ(๐ธ1)๐‘ƒ(๐ด|๐ธ1)+๐‘ƒ(๐ธ2)๐‘ƒ(๐ด|๐ธ2)+๐‘ƒ(๐ธ3)๐‘ƒ(๐ด|๐ธ3)+๐‘ƒ(๐ธ4)๐‘ƒ(๐ด|๐ธ4)) = (๐Ÿ/๐Ÿ’ ร— ๐Ÿ/๐Ÿ•)/( ๐Ÿ/๐Ÿ’ ร— ๐Ÿ‘/๐Ÿ๐Ÿ– + ๐Ÿ/๐Ÿ’ ร— ๐Ÿ/๐Ÿ’ + ๐Ÿ/๐Ÿ’ ร— ๐Ÿ/๐Ÿ• + ๐Ÿ/๐Ÿ’ ร— ๐Ÿ’/๐Ÿ๐Ÿ‘ ) = (1/28)/( 1/24 + 1/16 + 1/28 + 1/13 ) = (1/28)/( (1/(4 ร— 6) + 1/(4 ร— 4) + 1/(4 ร— 7)) + 1/13) = (1/28)/( ((4 ร— 7 + 6 ร— 7 + 6 ร— 4)/(4 ร— 6 ร— 4 ร— 7)) + 1/13) = (1/28)/((28 + 42 + 24)/(4 ร— 6 ร— 4 ร— 7) + 1/13) = (1/28)/(94/(4 ร— 6 ร— 4 ร— 7) + 1/13) = (1/28)/((94 ร— 13 + 4 ร— 6 ร— 4 ร— 7)/(13 ร— 4 ร— 6 ร— 4 ร— 7) ) = 1/((94 ร— 13 + 4 ร— 6 ร— 4 ร— 7)/(13 ร— 4 ร— 6) ) = (13 ร— 4 ร— 6)/( 94 ร— 13 + 4 ร— 6 ร— 4 ร— 7) = (13 ร— 2 ร— 6)/( 47 ร— 13 + 2 ร— 6 ร— 4 ร— 7) = 156/(611 + 336) = 156/947 = 0.164

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.