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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
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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 ? • 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﷯﷯ 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﷯ 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﷮15﷯ Selecting Box C = P(C|R) = 8﷮15﷯

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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