Misc 5 - Out of 100 students, two sections of 40, 60 are formed

Misc 5 - Chapter 16 Class 11 Probability - Part 2
Misc 5 - Chapter 16 Class 11 Probability - Part 3
Misc 5 - Chapter 16 Class 11 Probability - Part 4
Misc 5 - Chapter 16 Class 11 Probability - Part 5
Misc 5 - Chapter 16 Class 11 Probability - Part 6

Misc 5 - Chapter 16 Class 11 Probability - Part 7

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

Transcript

Misc 5 (a) Method 1 Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that you both enter the same sections? n(S) = 100C40 Let A be event that both enter section A n(A) = 98C38 n(S) = 100C60 Let B be event that both enter section B n(B) = 98C58 n(S) = 100C40, n(A) = 98C38 P(A) = (𝑛(𝐴))/(𝑛(𝑆)) = 𝟗𝟖𝑪𝟑𝟖/𝟏𝟎𝟎𝑪𝟒𝟎 = 98C38 ÷ 100C40 = 98!/38!(98 −38)! × 100!/40!(100 − 40)! = 98!/38!60! ÷ 100!/40!60! = 98!/38!60! × 40!60!/100! = (98! × 40! × 60!)/(38! × 60! × 100!) = (40 × 39)/(100 × 99) n(S) = 100C60, n(B) = 98C58 P(B) = (𝑛(𝐵))/(𝑛(𝑆)) = 𝟗𝟖𝑪𝟓𝟖/𝟏𝟎𝟎𝑪𝟔𝟎 = 98C58 ÷ 100C60 = 98!/58!(98 −58)! ÷ 100!/60!(100 −60)! = 98!/58!40! ÷ 100!/60!40! = 98!/58!40! × 60!40!/100! = (98! × 60! × 40! )/(58! × 40! ×100!) = (60 × 59)/(100 × 99) Probability that two students enter same section = Probability that both enter Section A + Probability that both enter Section B = P(A) + P(B) = (40 × 39)/(100 × 99) + (60 × 59)/(100 × 99) = 1560/9900 + 3540/9900 = 156/990 + 354/990 = (156 + 354)/990 = 510/990 = 51/99 = 𝟏𝟕/𝟑𝟑 Misc 5 (a) - Method 2 Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that you both enter the same sections? Both enter Section A It has 40 students Probability of selecting 2 students out of 40 from a total of 100 students = (40𝐶_2)/(100𝐶_2 ) Both enter Section B It has 60 students Probability of selecting 2 students out of 60 from a total of 100 students = (60𝐶_2)/(100𝐶_2 ) = ((40!/(2! 38!)))/((100!/(2! 98!)) ) = 40!/(2! 38!) ÷ 100!/(2! 98!) = 40!/(2! 38!) ÷ (2! 98!)/100! = (40 × 39)/(100 × 99) = ((60!/(2! 58!)))/((100!/(2! 98!)) ) = 60!/(2! 58!) ÷ 100!/(2! 98!) = 60!/(2! 58!) ÷ 2!98!/100! = (60 × 59)/(100 × 99) Thus, Probability that two students enter same section = Probability that both enter Section A + Probability that both enter Section B = P(A) + P(B) = (40 × 39)/(100 × 99) + (60 × 59)/(100 × 99) = 1560/9900 + 3540/9900 = 156/990 + 354/990 = (156 + 354)/990 = 510/990 = 51/99 = 𝟏𝟕/𝟑𝟑 Misc 5 Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that (b) you both enter the different sections? Probability that they enter different section = 1 – Probability that both enter in same section = 1 – 17/33 = (33 − 17)/33 = 𝟏𝟔/𝟑𝟑

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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.