Misc 12 - Suppose we have a four boxes A, B, C, D containing - Bayes theoram

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

Transcript

Misc 12 Suppose we have a four boxes A, B, C and D Containing colored marbles as given below : One of the Boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the Probability that it was drawn from box A? box B? box C ? Let R : Event that a red marble is drawn A : Event that the marble selected is from Box A B : Event that the marble selected is from Box B C : Event that the marble selected is from Box C D : Event that the marble selected is from Box D We need to find out Probability of drawn a marble is drawn from Box A, if it is a red. i.e. P(A|R) Also We need to find out Probability of drawn a marble is drawn from Box B, if it is a red. i.e. P(B|R) Also We need to find out Probability of drawn a marble is drawn from Box C, if it is a red. i.e. P(C|R) So, P(A|R) = (๐‘ƒ(๐ด). ๐‘ƒ(๐‘…|๐ด))/( ๐‘ƒ(๐ด). ๐‘ƒ(๐‘…|๐ด) + ๐‘ƒ(๐ต). ๐‘ƒ(๐‘…|๐ต) + ๐‘ƒ(๐ถ). ๐‘ƒ(๐‘…|๐ถ) + ๐‘ƒ(๐ท). ๐‘ƒ(๐‘…|๐ท) ) P(B|R) = (๐‘ƒ(๐ต). ๐‘ƒ(๐‘…|๐ต))/(๐‘ƒ(๐ด). ๐‘ƒ(๐‘…|๐ด) + ๐‘ƒ(๐ต). ๐‘ƒ(๐‘…|๐ต) + ๐‘ƒ(๐ถ). ๐‘ƒ(๐‘…|๐ถ) + ๐‘ƒ(๐ท). ๐‘ƒ(๐‘…|๐ท) ) P(C|R) = (๐‘ƒ(๐ถ). ๐‘ƒ(๐‘…|๐ถ))/(๐‘ƒ(๐ด). ๐‘ƒ(๐‘…|๐ด) + ๐‘ƒ(๐ต). ๐‘ƒ(๐‘…|๐ต) + ๐‘ƒ(๐ถ). ๐‘ƒ(๐‘…|๐ถ) + ๐‘ƒ(๐ท). ๐‘ƒ(๐‘…|๐ท) ) P(A) : Probability that Box A is selected (Since, Boxes are selected at random) = 1/4 P(R|A) = Probability that a red ball is selected from Box A = 1/10 P(B) : Probability that Box B is selected (Since, Boxes are selected at random) = 1/4 P(R|B) = Probability that a red ball is selected from Box B = 6/10 P(C) : Probability that Box C is selected (Since Boxes are selected at random) = 1/4 P(R|C) = Probability that a red ball is selected from Box C = 8/10 P(D) : Probability that Box D is selected (Since Boxes are selected at random) = 1/4 P(R|D) = Probability that a red ball is selected from Box D = 0/10 Putting value to Equations : P(A|R) = (๐‘ƒ(๐ด). ๐‘ƒ(๐‘…"|" ๐ด))/( ๐‘ƒ(๐ด). ๐‘ƒ(๐‘…"|" ๐ด) + ๐‘ƒ(๐ต). ๐‘ƒ(๐‘…"|" ๐ต) + ๐‘ƒ(๐ถ). ๐‘ƒ(๐‘…"|" ๐ถ) + ๐‘ƒ(๐ท). ๐‘ƒ(๐‘…"|" ๐ท) ) = (1/4 ร— 1/10)/( 1/4 ร— 1/10 + 1/4 ร— 6/10 + 1/4 ร— 8/10 + 1/4 ร— 0/10 ) = (1/4 ร— 1/10)/( 1/4 ร— 1/10 [1 + 6 + 8] ) = 1/15 P(B|R) = (๐‘ƒ(๐ต). ๐‘ƒ(๐‘…"|" ๐ต))/( ๐‘ƒ(๐ด). ๐‘ƒ(๐‘…"|" ๐ด) + ๐‘ƒ(๐ต). ๐‘ƒ(๐‘…"|" ๐ต) + ๐‘ƒ(๐ถ). ๐‘ƒ(๐‘…"|" ๐ถ) + ๐‘ƒ(๐ท). ๐‘ƒ(๐‘…"|" ๐ท) ) = (1/4 ร— 6/10)/( 1/4 ร— 1/10 + 1/4 ร— 6/10 + 1/4 ร— 8/10 + 1/4 ร— 0/10 ) = (1/4 ร— 1/10 ร— (6))/( 1/4 ร— 1/10 [1 + 6 + 8] ) = 6/15 = 2/5 P(C|R) = (๐‘ƒ(๐ถ). ๐‘ƒ(๐‘…"|" ๐ถ))/( ๐‘ƒ(๐ด). ๐‘ƒ(๐‘…"|" ๐ด) + ๐‘ƒ(๐ต). ๐‘ƒ(๐‘…"|" ๐ต) + ๐‘ƒ(๐ถ). ๐‘ƒ(๐‘…"|" ๐ถ) + ๐‘ƒ(๐ท). ๐‘ƒ(๐‘…"|" ๐ท) ) = (1/4 ร— 8/10)/( 1/4 ร— 1/10 + 1/4 ร— 6/10 + 1/4 ร— 8/10 + 1/4 ร— 0/10 ) = ((1/4 ร— 1/10) ร— (8))/((1/4 ร— 1/10) [1 + 6 + 8] ) = 8/15 Hence Probability when a red ball drawn : Selecting Box A = P(A|R) = 1/15 Selecting Box B = P(B|R) = 2/5 Selecting Box C = P(C|R) = 8/15

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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.