Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

Slide13.JPG

Slide14.JPG
Slide15.JPG

  1. Chapter 13 Class 12 Probability
  2. Serial order wise

Transcript

Example 21 A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.Let S1 : man speaks the truth S2 : man lies E : six on the die We need to find the Probability that it is actually a six, if the man reports that it a six i.e. P(S1|E) P(S1|E) = (๐‘ƒ(๐‘†_1 ).๐‘ƒ(๐ธ|๐‘†_1))/(๐‘ƒ(๐‘†_1 ).๐‘ƒ(๐ธ|๐‘†_1)+๐‘ƒ(๐‘†_2 ).๐‘ƒ(๐ธ|๐‘†_2)) P(S1) = Probability that man speaks truth = ๐Ÿ‘/๐Ÿ’ P(E|S1) = Probability that six appears on the die, if the man speaks the truth = ๐Ÿ/๐Ÿ” P(S2) = Probability man lies = 1 โ€“ P(E) = 1 โ€“ 3/4 = ๐Ÿ/๐Ÿ’ P(E|S2) = Probability that six appears on the die, if the man lies = P(6 does not appear) = 1 โ€“ 1/6 = ๐Ÿ“/๐Ÿ” Putting value in formula, P(S1|E) = (๐‘ƒ(๐‘†_1 ).๐‘ƒ(๐ธ|๐‘†_1))/(๐‘ƒ(๐‘†_1 ).๐‘ƒ(๐ธ|๐‘†_1)+๐‘ƒ(๐‘†_2 ).๐‘ƒ(๐ธ|๐‘†_2)) = (3/4 ร— 1/6)/( 3/4 ร— 1/6 + 1/4 ร— 5/6 ) = (1/4 ร— 1/6 ร— 3)/( 1/4 ร— 1/6 [3 + 5] ) = 3/8 Therefore, required probability is ๐Ÿ‘/๐Ÿ–

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.