Misc 3 - A die has two faces each with number '1', three - Miscellaneous

part 2 - Misc 3 - Miscellaneous - Serial order wise - Chapter 14 Class 11 Probability

part 3 - Misc 3 - Miscellaneous - Serial order wise - Chapter 14 Class 11 Probability

part 4 - Misc 3 - Miscellaneous - Serial order wise - Chapter 14 Class 11 Probability

part 5 - Misc 3 - Miscellaneous - Serial order wise - Chapter 14 Class 11 Probability part 6 - Misc 3 - Miscellaneous - Serial order wise - Chapter 14 Class 11 Probability

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Misc 3 A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine P(2) A normal die has 6 faces 1, 2, 3, 4, 5, 6 But in this question die has following 6 faces 1, 1, 2, 2, 2, 3 n(S) = 6 We need to find P(2) There are 3 three 2’s Hence, n(2) = 3 So, P(2) = 𝑛(2)/(𝑛(𝑆)) = 3/6 = 𝟏/𝟐 Misc 3 A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine (ii) P(1 or 3) Die has faces 1, 1, 2, 2, 2, 3 Now ,P (1 or 3) = P (1) + P(3) P(1) There are two 1’s Hence , n(1) = 2 P(1) = (𝑛(1))/(𝑛(𝑆)) = 2/6 P(3) There are only one 3’s Hence , n(3) = 1 P(3) = (𝑛(3))/(𝑛(𝑆)) = 1/6 Hence, P (1 or 3) = P (1) + P(3) = 2/6 + 1/6 = (2 + 1)/6 = 3/6 = 𝟏/𝟐 Misc 3 A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine (iii) P(not 3) P(not 3) = 1 – P(3) For P(3) There is only one 3’s Hence , n(3) = 1 P(3) = (𝑛(3))/(𝑛(𝑆)) = 1/6 Hence, P(not 3) = 1 – P(3) = 1 – 1/6 = (6 − 1)/6 = 𝟓/𝟔

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo