# Misc 12 - Chapter 13 Class 12 Probability

Last updated at Dec. 18, 2018 by Teachoo

Last updated at Dec. 18, 2018 by Teachoo

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(๐ด). ๐(๐ |๐ด))/( ๐(๐ด). ๐(๐ |๐ด) + ๐(๐ต). ๐(๐ |๐ต) + ๐(๐ถ). ๐(๐ |๐ถ) + ๐(๐ท). ๐(๐ |๐ท) ) P(B|R) = (๐(๐ต). ๐(๐ |๐ต))/(๐(๐ด). ๐(๐ |๐ด) + ๐(๐ต). ๐(๐ |๐ต) + ๐(๐ถ). ๐(๐ |๐ถ) + ๐(๐ท). ๐(๐ |๐ท) ) P(C|R) = (๐(๐ถ). ๐(๐ |๐ถ))/(๐(๐ด). ๐(๐ |๐ด) + ๐(๐ต). ๐(๐ |๐ต) + ๐(๐ถ). ๐(๐ |๐ถ) + ๐(๐ท). ๐(๐ |๐ท) ) 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(๐ด). ๐(๐ "|" ๐ด))/( ๐(๐ด). ๐(๐ "|" ๐ด) + ๐(๐ต). ๐(๐ "|" ๐ต) + ๐(๐ถ). ๐(๐ "|" ๐ถ) + ๐(๐ท). ๐(๐ "|" ๐ท) ) = (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(๐ต). ๐(๐ "|" ๐ต))/( ๐(๐ด). ๐(๐ "|" ๐ด) + ๐(๐ต). ๐(๐ "|" ๐ต) + ๐(๐ถ). ๐(๐ "|" ๐ถ) + ๐(๐ท). ๐(๐ "|" ๐ท) ) = (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(๐ถ). ๐(๐ "|" ๐ถ))/( ๐(๐ด). ๐(๐ "|" ๐ด) + ๐(๐ต). ๐(๐ "|" ๐ต) + ๐(๐ถ). ๐(๐ "|" ๐ถ) + ๐(๐ท). ๐(๐ "|" ๐ท) ) = (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

Miscellaneous

Misc 1

Misc 2 Important

Misc 3

Misc 4

Misc 5 Important Deleted for CBSE Board 2022 Exams

Misc 6 Important Deleted for CBSE Board 2022 Exams

Misc 7 Important

Misc 8 Important

Misc 9 Important Deleted for CBSE Board 2022 Exams

Misc 10 Important

Misc 11 Important

Misc 12 You are here

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18

Misc 19

Chapter 13 Class 12 Probability (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.