Probability that "a" speaks truth is 4/5. A coin is tossed. A reports that a head appears. Find out the probability that actually there was head?

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Madhur Jain

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4/5 by Bayes theorm
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E = event that 'A' reports it is head. E1 = The outcome is head. E2 = outcome is tail. Required. P(E1/E) = P(E/E1) * P(E1)/ P(E) Cases in which 'A' reports heads case 1 when it is head and he is saying the truth = 1/2 * 4/5 case 2 when it is a tail and he is lying. = 1/2 *1/5. Also...
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E = event that 'A' reports it is head. E1 = The outcome is head. E2 = outcome is tail. Required. P(E1/E) = P(E/E1) * P(E1)/ P(E) Cases in which 'A' reports heads case 1 when it is head and he is saying the truth = 1/2 * 4/5 case 2 when it is a tail and he is lying. = 1/2 *1/5. Also P(E/E1) = 4/5 (Since, it just denotes the probability of 'A' telling the truth.) So, P(E1/E) = (4/5) * (1/2)/ [{(4/5) * (1/2)}+{(1/5) * (1/2)}] = 4/5 read less
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Maths Teacher

Let E1 be the event A speaks ruth and E2 A does not. A is the event a headappears By Baye's theorem P(E1/A ) = 4/10 /5/10= 4/5
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Mathematics

The correct answer is 4/5
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Let E1 be the event A speaks ruth and E2 A does not. A is the event a headappears By Baye's theorem P(E1/A ) = 4/10 /5/10= 4/5
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Trainer

The probability that there was a head= A speaks truth x probability of falling head= (4/5)x(1/2)= 2/5 ( provided the coin is tossed one time only)
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