Example 7 - Chapter 13 Class 12 Probability (Important Question)
Last updated at Feb. 15, 2020 by Teachoo
Chapter 13 Class 12 Probability
Example 6
Ex 13.1, 10 (a) Important
Ex 13.1, 12 Important
Example 11 Important
Ex 13.2, 7 Important
Ex 13.2, 11 (i)
Ex 13.2, 14 Important
Example 17 Important
Example 18 Important
Example 20 Important
Example 21 Important
Ex 13.3, 2 Important
Ex 13.3, 4 Important
Ex 13.3, 8 Important
Ex 13.3, 10 Important
Ex 13.3, 12 Important
Ex 13.3, 13 (MCQ) Important
Example 25 Important
Example 26 Important
Example 27 Deleted for CBSE Board 2022 Exams
Example 28 Important Deleted for CBSE Board 2022 Exams
Example 29 Important Deleted for CBSE Board 2022 Exams
Ex 13.4, 3 Important
Ex 13.4, 6 Important
Ex 13.4, 11 Important Deleted for CBSE Board 2022 Exams
Ex 13.4, 15 Deleted for CBSE Board 2022 Exams
Example 31 Important Deleted for CBSE Board 2022 Exams
Example 32 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 4 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 6 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 10 Important Deleted for CBSE Board 2022 Exams
Ex 13.5, 13 Important Deleted for CBSE Board 2022 Exams
Example 35 Deleted for CBSE Board 2022 Exams
Example 36 Important
Misc 6 Important Deleted for CBSE Board 2022 Exams
Misc 9 Deleted for CBSE Board 2022 Exams
Misc 11 Important
Misc 13 Important
Misc 16 Important
Chapter 13 Class 12 Probability
Last updated at Feb. 15, 2020 by Teachoo
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