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Ex 14.2 ,5 - Chapter 14 Class 11 Probability
Last updated at May 29, 2023 by Teachoo
Ex 14.2
Ex 14.2, 1
Ex 14.2, 2
Ex 14.2, 3 (i) Important
Ex 14.2, 3 (ii)
Ex 14.2, 3 (iii)
Ex 14.2, 3 (iv)
Ex 14.2, 3 (v)
Ex 14.2, 4 Important
Ex 14.2 ,5 Important You are here
Ex 14.2, 6
Ex 14.2, 7 Important
Ex 14.2, 8 Important
Ex 14.2, 9
Ex 14.2, 10
Ex 14.2, 11 Important
Ex 14.2, 12 (i)
Ex 14.2, 12 (ii) Important
Ex 14.2, 13
Ex 14.2, 14 Important
Ex 14.2, 15 Important
Ex 14.2, 16 Important
Ex 14.2, 17
Ex 14.2, 18
Ex 14.2, 19
Ex 14.2, 20 Important
Ex 14.2, 21 Important
Chapter 14 Class 11 Probability
Serial order wise
- Ex 14.1
-
Ex 14.2
- Examples
- Miscellaneous
- Sample Space
Transcript
Ex 14.2, 5 A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 If the coin is tossed we get only 1 or 6 If a die is thrown we get 1, 2, 3, 4, 5, 6 Hence, S = {█("(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)," @" (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)" )} n(S) = 12 Let A be the event sum 3 Hence A = {(1, 2)} n(A) = 1 Probability of getting sum as 3 = P(A) = (n(A))/(n(S)) = 𝟏/𝟏𝟐 Ex 14.2, 5 A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (ii) 12 Let B be the event sum is 12. Hence B = {(6, 6)} n(B) = 1 Probability of getting sum as 12 = P(B) = (n(B))/(n(S)) = 𝟏/𝟏𝟐
