


Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Examples
Example 2
Example 3
Example 4
Example 5 Important
Example 6
Example 7 Important You are here
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
Example 23
Example 24 Important
Example 25 Important
Example 26 Important
Example 27
Example 28 Important Deleted for CBSE Board 2023 Exams
Example 29 Important
Example 30 Important Deleted for CBSE Board 2023 Exams
Example 31 Important Deleted for CBSE Board 2023 Exams
Example 32 Important Deleted for CBSE Board 2023 Exams
Example 33 Important
Example 34 Deleted for CBSE Board 2023 Exams
Example 35
Example 36 Important
Example 37 Important
Last updated at Feb. 15, 2020 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Example 7 Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail’.A coin is tossed If the coin shows head, it is tossed again. If it shows tail, then a die is thrown. Hence different value of probabilities are We need to find the probability that the die shows a number greater than 4, given that there is at least one tail. Now, E : at least one tail F : Number greater than 4 on the die We need to find P(E|F) Also, E ∩ F = {(T, 5), (T, 6)} E = { (T, 5), (T, 6) } P(E) = P(T, 5), P(T, 6) = 1/12 + 1/12 = 2/12 = 1/6 F = { (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) } P(F) = P(H,T) + P(T,1) + P(T,2) + P(T,3) + P(T,4) + P(T,5) + P(T,6) = 1/4 + 1/12 + 1/12 +1/12 +1/12 +1/12 + 1/12 + 1/12 = 1/4 + 6/12 = 1/4 + 1/2 = 3/4 So, P(E ∩ F) = P(T, 5) + P(T, 6) = 1/12 + 1/12 = 2/12 = 1/6 Now, P(E|F) = (𝑃(𝐸 ∩ 𝐹))/(𝑃(𝐹)) = (1/6)/(3/4) = 1/6 × 4/3 = 𝟐/𝟗 Therefore, required probability is 2/9