Misc 6 - Chapter 13 Class 12 Probability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Misc 6 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
Miscellaneous
Misc 1 (ii)
Misc 2 (i) Important
Misc 2 (ii)
Misc 3
Misc 4
Misc 5 Important
Misc 6 You are here
Misc 7 Important
Misc 8 Important
Misc 9 Important
Misc 10 Important
Misc 11 (MCQ) Important
Misc 12 (MCQ)
Misc 13 (MCQ)
Question 1 Important
Question 2 Important
Question 3 Important
Question 4
Question 5 Important
Question 6 Important
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo