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

Misc 12 - Chapter 13 Class 12 Probability - Part 2
Misc 12 - Chapter 13 Class 12 Probability - Part 3
Misc 12 - Chapter 13 Class 12 Probability - Part 4
Misc 12 - Chapter 13 Class 12 Probability - Part 5
Misc 12 - Chapter 13 Class 12 Probability - Part 6

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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) = (𝑃(𝐴). 𝑃(𝑅|𝐴))/( 𝑃(𝐴). 𝑃(𝑅|𝐴) + 𝑃(𝐵). 𝑃(𝑅|𝐵) + 𝑃(𝐶). 𝑃(𝑅|𝐶) + 𝑃(𝐷). 𝑃(𝑅|𝐷) ) P(B|R) = (𝑃(𝐵). 𝑃(𝑅|𝐵))/(𝑃(𝐴). 𝑃(𝑅|𝐴) + 𝑃(𝐵). 𝑃(𝑅|𝐵) + 𝑃(𝐶). 𝑃(𝑅|𝐶) + 𝑃(𝐷). 𝑃(𝑅|𝐷) ) P(C|R) = (𝑃(𝐶). 𝑃(𝑅|𝐶))/(𝑃(𝐴). 𝑃(𝑅|𝐴) + 𝑃(𝐵). 𝑃(𝑅|𝐵) + 𝑃(𝐶). 𝑃(𝑅|𝐶) + 𝑃(𝐷). 𝑃(𝑅|𝐷) ) 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) = (𝑃(𝐴). 𝑃(𝑅"|" 𝐴))/( 𝑃(𝐴). 𝑃(𝑅"|" 𝐴) + 𝑃(𝐵). 𝑃(𝑅"|" 𝐵) + 𝑃(𝐶). 𝑃(𝑅"|" 𝐶) + 𝑃(𝐷). 𝑃(𝑅"|" 𝐷) ) = (1/4 × 1/10)/( 1/4 × 1/10 + 1/4 × 6/10 + 1/4 × 8/10 + 1/4 × 0/10 ) = (1/4 × 1/10)/( 1/4 × 1/10 [1 + 6 + 8] ) = 1/15 P(B|R) = (𝑃(𝐵). 𝑃(𝑅"|" 𝐵))/( 𝑃(𝐴). 𝑃(𝑅"|" 𝐴) + 𝑃(𝐵). 𝑃(𝑅"|" 𝐵) + 𝑃(𝐶). 𝑃(𝑅"|" 𝐶) + 𝑃(𝐷). 𝑃(𝑅"|" 𝐷) ) = (1/4 × 6/10)/( 1/4 × 1/10 + 1/4 × 6/10 + 1/4 × 8/10 + 1/4 × 0/10 ) = (1/4 × 1/10 × (6))/( 1/4 × 1/10 [1 + 6 + 8] ) = 6/15 = 2/5 P(C|R) = (𝑃(𝐶). 𝑃(𝑅"|" 𝐶))/( 𝑃(𝐴). 𝑃(𝑅"|" 𝐴) + 𝑃(𝐵). 𝑃(𝑅"|" 𝐵) + 𝑃(𝐶). 𝑃(𝑅"|" 𝐶) + 𝑃(𝐷). 𝑃(𝑅"|" 𝐷) ) = (1/4 × 8/10)/( 1/4 × 1/10 + 1/4 × 6/10 + 1/4 × 8/10 + 1/4 × 0/10 ) = ((1/4 × 1/10) × (8))/((1/4 × 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.