Example 33 - Chapter 13 Class 12 Probability (Term 2)
Last updated at Feb. 15, 2020 by Teachoo
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Example 33 Important You are here
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Example 36 Important
Example 37 Important
Chapter 13 Class 12 Probability (Term 2)
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Transcript
Example 33 Coloured balls are distributed in four boxes as shown in the following table: A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the boxs III?Let A : Event that a black ball is selected E1 : Event that the ball is selected from box I E2 : Event that the ball is selected from box II E3 : Event that the ball is selected from box III E4 : Event that the ball is selected from box IV We need to find out the Probability of the ball drawn is from box III if it is black. i.e. P(E3|A) P(E3|A)= (π(πΈ3).π(π΄|πΈ3))/(π(πΈ1)π(π΄|πΈ1)+π(πΈ2)π(π΄|πΈ2)+π(πΈ3)π(π΄|πΈ3)+π(πΈ4)π(π΄|πΈ4)) P(E1) = Probability that ball drawn is from box I = 1/4 P(A|E1) = Probability of that black ball is selected from Box I = (ππ’ππππ ππ πππππ )/(πππ‘ππ ππ’ππππ ππ πππππ ππ π‘βπ πππ₯) = 3/18 P(E2) = Probability that ball drawn is from box II = 1/4 P(A|E2) = Probability of that black ball is selected from Box II = (ππ’ππππ ππ πππππ )/(πππ‘ππ ππ’ππππ ππ πππππ ππ π‘βπ πππ₯) = 2/8 = 1/4 P(E3) = Probability that ball drawn is from box III = 1/4 P(A|E3) = Probability of that black ball is selected froAm Box II = (ππ’ππππ ππ πππππ )/(πππ‘ππ ππ’ππππ ππ πππππ ππ π‘βπ πππ₯) = 1/7 P(E4) = Probability that ball drawn is from box IV = 1/4 P(A|E4) = Probability of that black ball is selected from Box IV = (ππ’ππππ ππ πππππ )/(πππ‘ππ ππ’ππππ ππ πππππ ππ π‘βπ πππ₯) = 4/13 Putting values in Equation "P(E3|A)"=(π(πΈ3).π(π΄|πΈ3))/(π(πΈ1)π(π΄|πΈ1)+π(πΈ2)π(π΄|πΈ2)+π(πΈ3)π(π΄|πΈ3)+π(πΈ4)π(π΄|πΈ4)) = (1/4 Γ 1/7)/( 1/4 Γ 3/18 + 1/4 Γ 1/4 + 1/4 Γ 1/7 + 1/4 Γ 4/13 ) = (1/28)/( 1/24 + 1/16 + 1/28 + 1/13 ) = (1/28)/( (1/(4 Γ 6) + 1/(4 Γ 4) + 1/(4 Γ 7)) + 1/13) = (1/28)/( ((4 Γ 7 + 6 Γ 7 + 6 Γ 4)/(4 Γ 6 Γ 4 Γ 7)) + 1/13) = (1/28)/((28 + 42 + 24)/(4 Γ 6 Γ 4 Γ 7) + 1/13) = (1/28)/(94/(4 Γ 6 Γ 4 Γ 7) + 1/13) = (1/28)/((94 Γ 13 + 4 Γ 6 Γ 4 Γ 7)/(13 Γ 4 Γ 6 Γ 4 Γ 7) ) = 1/((94 Γ 13 + 4 Γ 6 Γ 4 Γ 7)/(13 Γ 4 Γ 6) ) = (13 Γ 4 Γ 6)/( 94 Γ 13 + 4 Γ 6 Γ 4 Γ 7) = (13 Γ 2 Γ 6)/( 47 Γ 13 + 2 Γ 6 Γ 4 Γ 7) = 156/(611 + 336) = 156/947 = 0.164
