@Shubhodeep “Other side can have a head if and only if the picked coin is C3” if coin is C2 then also the other side will be head , why u have not include the possibility for coin C2 ? plz explain

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Suppose that Amitabh Bachchan has ten coins in his pocket. 3 coins have tails on both sides. 4 coins have heads on both sides. 3 coins have heads on one side and tails on the other and both the outcomes are equally likely when that coin is flipped. In a bet with Dharmendra, Amitabh picks up a coin at random (each coin is equally likely to be picked) from these ten coins, flips it and finds that the outcome is tails. What is the probability that the other side of this coin is heads?

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Best answer

Lets assume

C1: coin having both side Tails

C2: coin having both side Head

C3: coin having one side Head and other side tail

$P(C1) = 3/10$

$P(C2) = 4/10$

$P(C3) = 3/10$

$P(Tails) = P(C1).P(Tails|C1) + P(C2).P(Tails|C2) + P(C3).P(Tails|C3)$

$= (3/10)(2/2) + (4/10)(0/2) + (3/10)(1/2)$

Other side can have a head if and only if the picked coin is C3

we can also frame the question as find the probability that tails was from coin C3

So required probability = $\frac{(3/10)(1/2)}{(3/10)(2/2) + (4/10)(0/2) + (3/10)(1/2)}$ = $\frac{1}{3}$

C1: coin having both side Tails

C2: coin having both side Head

C3: coin having one side Head and other side tail

$P(C1) = 3/10$

$P(C2) = 4/10$

$P(C3) = 3/10$

$P(Tails) = P(C1).P(Tails|C1) + P(C2).P(Tails|C2) + P(C3).P(Tails|C3)$

$= (3/10)(2/2) + (4/10)(0/2) + (3/10)(1/2)$

Other side can have a head if and only if the picked coin is C3

we can also frame the question as find the probability that tails was from coin C3

So required probability = $\frac{(3/10)(1/2)}{(3/10)(2/2) + (4/10)(0/2) + (3/10)(1/2)}$ = $\frac{1}{3}$

@Shubhodeep “Other side can have a head if and only if the picked coin is C3” if coin is C2 then also the other side will be head , why u have not include the possibility for coin C2 ? plz explain

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