Check sibling questions

Example 33 - Coloured balls are distributed in four boxes

Example 33 - Chapter 13 Class 12 Probability - Part 2
Example 33 - Chapter 13 Class 12 Probability - Part 3
Example 33 - Chapter 13 Class 12 Probability - Part 4
Example 33 - Chapter 13 Class 12 Probability - Part 5


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)= (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 = 1/4 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‘₯) = 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 = (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 = 1/4 P(A|E3) = Probability of that black ball is selected froAm 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‘₯) = 1/7 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 = (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‘₯) = 4/13 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)) = (1/4 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

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.