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

Misc 3 - Chapter 16 Class 11 Probability - Part 2

Misc 3 - Chapter 16 Class 11 Probability - Part 3

Misc 3 - Chapter 16 Class 11 Probability - Part 4

Misc 3 - Chapter 16 Class 11 Probability - Part 5

Misc 3 - Chapter 16 Class 11 Probability - Part 6

  1. Chapter 16 Class 11 Probability (Term 2)
  2. Serial order wise

Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.