Example 22 - Chapter 13 Class 12 Probability

Last updated at August 19, 2026 by

Example 22 - Coloured balls are distributed in four boxes - Examples

part 2 - Example 22 - Examples - Serial order wise - Chapter 13 Class 12 Probability
part 3 - Example 22 - Examples - Serial order wise - Chapter 13 Class 12 Probability part 4 - Example 22 - Examples - Serial order wise - Chapter 13 Class 12 Probability part 5 - Example 22 - Examples - Serial order wise - Chapter 13 Class 12 Probability

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Transcript

Example 22 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)= (𝑃(𝐸3).𝑃(𝐴|𝐸3))/(𝑃(𝐸1)𝑃(𝐴|𝐸1)+𝑃(𝐸2)𝑃(𝐴|𝐸2)+𝑃(𝐸3)𝑃(𝐴|𝐸3)+𝑃(𝐸4)𝑃(𝐴|𝐸4)) P(E1) = Probability that ball drawn is from box I = 𝟏/πŸ’ P(A|E1) = Probability of that black ball is selected from Box I = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘™π‘™π‘ )/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘™π‘™π‘  𝑖𝑛 π‘‘β„Žπ‘’ π‘π‘œπ‘₯) = πŸ‘/πŸπŸ– P(E2) = Probability that ball drawn is from box II = 𝟏/πŸ’ P(A|E2) = Probability of that black ball is selected from Box II = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘™π‘™π‘ )/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘™π‘™π‘  𝑖𝑛 π‘‘β„Žπ‘’ π‘π‘œπ‘₯) = 2/8 = 1/4 P(E3) = Probability that ball drawn is from box III = 𝟏/πŸ’ P(A|E3) = Probability of that black ball is selected from Box II = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘™π‘™π‘ )/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘™π‘™π‘  𝑖𝑛 π‘‘β„Žπ‘’ π‘π‘œπ‘₯) = 𝟏/πŸ• P(E4) = Probability that ball drawn is from box IV = 𝟏/πŸ’ P(A|E4) = Probability of that black ball is selected from Box IV = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘™π‘™π‘ )/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘™π‘™π‘  𝑖𝑛 π‘‘β„Žπ‘’ π‘π‘œπ‘₯) = πŸ’/πŸπŸ‘ Putting values in Equation "P(E3|A)"=(𝑃(𝐸3).𝑃(𝐴|𝐸3))/(𝑃(𝐸1)𝑃(𝐴|𝐸1)+𝑃(𝐸2)𝑃(𝐴|𝐸2)+𝑃(𝐸3)𝑃(𝐴|𝐸3)+𝑃(𝐸4)𝑃(𝐴|𝐸4)) = (𝟏/πŸ’ Γ— 𝟏/πŸ•)/( 𝟏/πŸ’ Γ— πŸ‘/πŸπŸ– + 𝟏/πŸ’ Γ— 𝟏/πŸ’ + 𝟏/πŸ’ Γ— 𝟏/πŸ• + 𝟏/πŸ’ Γ— πŸ’/πŸπŸ‘ ) = (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

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