# Ex 13.2, 12 - Chapter 13 Class 12 Probability

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 13.2, 12 A die is tossed thrice. Find the Probability of getting an odd number at least once. A die is tossed thrice Probability of getting an odd number at least once = Probability of getting an odd number once + Probability of getting an odd number twice + Probability of getting an odd number thrice = 1 Probability of getting an odd number 0 times = 1 Probability of getting an even number all 3 times Let the events be O : getting an odd number E : getting an even number Now, Required Probability = 1 P(EEE) So, P(EEE) = P(E) P(E|E) P(EE|E) P(E) = Probability of getting an even number Even numbers on a die = {2, 4, 6} = 3 Total numbers on a die = {1, 2, , 6} = 6 P(E) = 3 6 = 1 2 P(E|E) = Probability of getting an even number on the second die given even number appeared on the first die Even numbers on a die = {2, 4, 6} = 3 Total numbers on a die = {1, 2, , 6} = 6 P(E|E) = 3 6 = 1 2 P(EE|E) = Probability of getting an even number on the third die, given even numbers appeared on the first two die. Even numbers on a die = {2, 4, 6} = 3 Total numbers on a die = {1, 2, , 6} = 6 P(EE|E) = 3 6 = 1 2 So, P(EEE) = P(E) P(E|E) P(EE|E) P(EEE) = 1 2 1 2 1 2 = 1 8 Probability of getting an odd number at least once = 1 P(EEE) = 1 1 8 =

Chapter 13 Class 12 Probability

Serial order wise

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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.