Example 33 - Coloured balls are distributed in four boxes - Examples

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

Transcript

Example 33 Coloured balls are distributed in four boxes as shown in the following table: Box Colour Black White Red Blue I 3 4 5 6 II 2 2 2 2 III 1 2 3 1 IV 4 3 1 5 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 box 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 = 1/4 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) = 3/18 P(E3) = Probability that ball drawn is from box III = 1/4 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) = 1/7 P(E2) = Probability that ball drawn is from box II = 1/4 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(E4) = Probability that ball drawn is from box IV = 1/4 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) = 4/13 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)) = (1/4 ร— 1/7)/( 1/4 ร— 3/8 + 1/4 ร— 1/4 + 1/4 ร— 1/7 + 1/4 ร— 4/13 ) = (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 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.