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A man is known to speak truth 3 out of 4 times. He throws a die and

Example 21 - Chapter 13 Class 12 Probability - Part 2
Example 21 - Chapter 13 Class 12 Probability - Part 3


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) = (𝑃(𝑆_1 ).𝑃(𝐸|𝑆_1))/(𝑃(𝑆_1 ).𝑃(𝐸|𝑆_1)+𝑃(𝑆_2 ).𝑃(𝐸|𝑆_2)) P(S1) = Probability that man speaks truth = πŸ‘/πŸ’ P(E|S1) = Probability that six appears on the die, if the man speaks the truth = 𝟏/πŸ” P(S2) = Probability man lies = 1 – P(E) = 1 – 3/4 = 𝟏/πŸ’ P(E|S2) = Probability that six appears on the die, if the man lies = P(6 does not appear) = 1 – 1/6 = πŸ“/πŸ” Putting value in formula, P(S1|E) = (𝑃(𝑆_1 ).𝑃(𝐸|𝑆_1))/(𝑃(𝑆_1 ).𝑃(𝐸|𝑆_1)+𝑃(𝑆_2 ).𝑃(𝐸|𝑆_2)) = (3/4 Γ— 1/6)/( 3/4 Γ— 1/6 + 1/4 Γ— 5/6 ) = (1/4 Γ— 1/6 Γ— 3)/( 1/4 Γ— 1/6 [3 + 5] ) = 3/8 Therefore, required probability is πŸ‘/π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.