# Ex 13.1, 6 - Chapter 13 Class 12 Probability

Last updated at June 29, 2018 by Teachoo

Last updated at June 29, 2018 by Teachoo

Transcript

Ex 13.1, 6 A coin is tossed three times, where (i) E : head on third toss , F : heads on first two tosses Coin is tossed three times S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} E : head on third toss E = {HHH, HTH, THH, TTH} P(E) = 4/8=1/2 F : heads on first two tosses F = {HHH, HHT} P(F) = 2/8=1/4 Also, E F = {HHH} P (E F ) = 1/8 We need to find P(E|F) P(E|F) = ( ( ))/( ( )) = (1/8 )/(1/4 ) = 1/8 4/1 = 1/2 P(E|F) = / Ex 13.1, 6 A coin is tossed three times, where (ii) E : at least two heads , F : at most two heads Coin is tossed three times S = {HHH, HHT, THH, HTH, TTH, THT, HTT, TTT} E : at least two heads E = {HHT, THH, HTH, HHH} P(E) = 4/8=1/2 F : at most two heads F = {HHT, THH, HTH, TTH, THT, HTT, TTT} P(F) = 7/8 Also, E F = {HHT, THH, HTH} P(E F ) = 3/8 Now, P(E|F) = ( ( ))/( ( )) = " " ( 3/8 )/(7/8 ) = 3/8 8/7 = 3/7 P(E|F) = / Ex 13.1, 6 A coin is tossed three times, where (iii) E : at most two tails , F : at least one tail A coin is tossed 3 times S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} E : at most two tails E = {HHH, HHT, HTH, THH, TTH, THT, HHT} P(E) = 7/8 F : at least one tail F = {HHT, HTH, THH, TTH THT, HTT, TTT} P(F) = 7/8 Also, E F = {HHT, HTH, THH, TTH, THT, HTT} P(E F) = 6/8 Now, P(E|F) = ( ( ))/( ( )) = (6/8 )/(7/8 ) = 6/8 8/7 = 6/7 P(E|F) = /

Chapter 13 Class 12 Probability

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.