Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Examples

Example 1

Example 2

Example 3

Example 4

Example 5 Important

Example 6

Example 7 Important

Example 8

Example 9 Important

Example 10

Example 11 Important

Example 12 Important

Example 13 Important

Example 14 Important

Example 15 Important

Example 16

Example 17 Important

Example 18 Important

Example 19 Important

Example 20 Important

Example 21 Important

Example 22 Important You are here

Example 23 Important

Example 24 Important

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 Important Deleted for CBSE Board 2024 Exams

Question 4 Important Deleted for CBSE Board 2024 Exams

Question 5 Important Deleted for CBSE Board 2024 Exams

Question 6 Deleted for CBSE Board 2024 Exams

Question 7 Important Deleted for CBSE Board 2024 Exams

Question 8 Important Deleted for CBSE Board 2024 Exams

Question 9 Important Deleted for CBSE Board 2024 Exams

Question 10 Important Deleted for CBSE Board 2024 Exams

Question 11 Important Deleted for CBSE Board 2024 Exams

Question 12 Deleted for CBSE Board 2024 Exams

Question 13 Deleted for CBSE Board 2024 Exams

Chapter 13 Class 12 Probability

Serial order wise

Last updated at Aug. 16, 2023 by Teachoo

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 = 𝟏/𝟒 P(A|E1) = Probability of that black ball is selected from Box I = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑎𝑙𝑙𝑠)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑎𝑙𝑙𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑜𝑥) = 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 = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑎𝑙𝑙𝑠)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑎𝑙𝑙𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑜𝑥) = 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 from Box II = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑎𝑙𝑙𝑠)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑎𝑙𝑙𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑜𝑥) = 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 = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑎𝑙𝑙𝑠)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑎𝑙𝑙𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑜𝑥) = 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 × 𝟏/𝟕)/( 𝟏/𝟒 × 𝟑/𝟏𝟖 + 𝟏/𝟒 × 𝟏/𝟒 + 𝟏/𝟒 × 𝟏/𝟕 + 𝟏/𝟒 × 𝟒/𝟏𝟑 ) = (1/28)/( 1/24 + 1/16 + 1/28 + 1/13 ) = (1/28)/( (1/(4 × 6) + 1/(4 × 4) + 1/(4 × 7)) + 1/13) = (1/28)/( ((4 × 7 + 6 × 7 + 6 × 4)/(4 × 6 × 4 × 7)) + 1/13) = (1/28)/((28 + 42 + 24)/(4 × 6 × 4 × 7) + 1/13) = (1/28)/(94/(4 × 6 × 4 × 7) + 1/13) = (1/28)/((94 × 13 + 4 × 6 × 4 × 7)/(13 × 4 × 6 × 4 × 7) ) = 1/((94 × 13 + 4 × 6 × 4 × 7)/(13 × 4 × 6) ) = (13 × 4 × 6)/( 94 × 13 + 4 × 6 × 4 × 7) = (13 × 2 × 6)/( 47 × 13 + 2 × 6 × 4 × 7) = 156/(611 + 336) = 156/947 = 0.164