Bayes theorem

Chapter 13 Class 12 Probability (Term 2)
Concept wise

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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) = (π(π΄). π(π|π΄))/( π(π΄). π(π|π΄) + π(π΅). π(π|π΅) + π(πΆ). π(π|πΆ) + π(π·). π(π|π·) ) P(B|R) = (π(π΅). π(π|π΅))/(π(π΄). π(π|π΄) + π(π΅). π(π|π΅) + π(πΆ). π(π|πΆ) + π(π·). π(π|π·) ) P(C|R) = (π(πΆ). π(π|πΆ))/(π(π΄). π(π|π΄) + π(π΅). π(π|π΅) + π(πΆ). π(π|πΆ) + π(π·). π(π|π·) ) 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) = (π(π΄). π(π"|" π΄))/( π(π΄). π(π"|" π΄) + π(π΅). π(π"|" π΅) + π(πΆ). π(π"|" πΆ) + π(π·). π(π"|" π·) ) = (1/4 Γ 1/10)/( 1/4 Γ 1/10 + 1/4 Γ 6/10 + 1/4 Γ 8/10 + 1/4 Γ 0/10 ) = (1/4 Γ 1/10)/( 1/4 Γ 1/10 [1 + 6 + 8] ) = 1/15 P(B|R) = (π(π΅). π(π"|" π΅))/( π(π΄). π(π"|" π΄) + π(π΅). π(π"|" π΅) + π(πΆ). π(π"|" πΆ) + π(π·). π(π"|" π·) ) = (1/4 Γ 6/10)/( 1/4 Γ 1/10 + 1/4 Γ 6/10 + 1/4 Γ 8/10 + 1/4 Γ 0/10 ) = (1/4 Γ 1/10 Γ (6))/( 1/4 Γ 1/10 [1 + 6 + 8] ) = 6/15 = 2/5 P(C|R) = (π(πΆ). π(π"|" πΆ))/( π(π΄). π(π"|" π΄) + π(π΅). π(π"|" π΅) + π(πΆ). π(π"|" πΆ) + π(π·). π(π"|" π·) ) = (1/4 Γ 8/10)/( 1/4 Γ 1/10 + 1/4 Γ 6/10 + 1/4 Γ 8/10 + 1/4 Γ 0/10 ) = ((1/4 Γ 1/10) Γ (8))/((1/4 Γ 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