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

Misc 5 - Chapter 16 Class 11 Probability - Part 2
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Misc 5 - Chapter 16 Class 11 Probability - Part 7

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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) = (๐‘›(๐ด))/(๐‘›(๐‘†)) = ๐Ÿ—๐Ÿ–๐‘ช๐Ÿ‘๐Ÿ–/๐Ÿ๐ŸŽ๐ŸŽ๐‘ช๐Ÿ’๐ŸŽ = 98C38 รท 100C40 = 98!/38!(98 โˆ’38)! ร— 100!/40!(100 โˆ’ 40)! = 98!/38!60! รท 100!/40!60! = 98!/38!60! ร— 40!60!/100! = (98! ร— 40! ร— 60!)/(38! ร— 60! ร— 100!) = (40 ร— 39)/(100 ร— 99) n(S) = 100C60, n(B) = 98C58 P(B) = (๐‘›(๐ต))/(๐‘›(๐‘†)) = ๐Ÿ—๐Ÿ–๐‘ช๐Ÿ“๐Ÿ–/๐Ÿ๐ŸŽ๐ŸŽ๐‘ช๐Ÿ”๐ŸŽ = 98C58 รท 100C60 = 98!/58!(98 โˆ’58)! รท 100!/60!(100 โˆ’60)! = 98!/58!40! รท 100!/60!40! = 98!/58!40! ร— 60!40!/100! = (98! ร— 60! ร— 40! )/(58! ร— 40! ร—100!) = (60 ร— 59)/(100 ร— 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 ร— 39)/(100 ร— 99) + (60 ร— 59)/(100 ร— 99) = 1560/9900 + 3540/9900 = 156/990 + 354/990 = (156 + 354)/990 = 510/990 = 51/99 = ๐Ÿ๐Ÿ•/๐Ÿ‘๐Ÿ‘ 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 = (40๐ถ_2)/(100๐ถ_2 ) Both enter Section B It has 60 students Probability of selecting 2 students out of 60 from a total of 100 students = (60๐ถ_2)/(100๐ถ_2 ) = ((40!/(2! 38!)))/((100!/(2! 98!)) ) = 40!/(2! 38!) รท 100!/(2! 98!) = 40!/(2! 38!) รท (2! 98!)/100! = (40 ร— 39)/(100 ร— 99) = ((60!/(2! 58!)))/((100!/(2! 98!)) ) = 60!/(2! 58!) รท 100!/(2! 98!) = 60!/(2! 58!) รท 2!98!/100! = (60 ร— 59)/(100 ร— 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 ร— 39)/(100 ร— 99) + (60 ร— 59)/(100 ร— 99) = 1560/9900 + 3540/9900 = 156/990 + 354/990 = (156 + 354)/990 = 510/990 = 51/99 = ๐Ÿ๐Ÿ•/๐Ÿ‘๐Ÿ‘ 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 โ€“ Probability that both enter in same section = 1 โ€“ 17/33 = (33 โˆ’ 17)/33 = ๐Ÿ๐Ÿ”/๐Ÿ‘๐Ÿ‘

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.