GATE2014 EC-1: GA-10

Question

You are given three coins: one has heads on both faces, the second has tails on both faces, and the third has a head on one face and a tail on the other. You choose a coin at random and toss it, and it comes up heads. The probability that the other face is tails is

• A. $\dfrac{1}{4}$ 
• B. $\dfrac{1}{3}$ 
• C. $\dfrac{1}{2}$ 
• D. $\dfrac{2}{3}$ 

Comments

This answer is more logical

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@smsubham @Abhrajyoti00

I am not able to understand how

$E_{1} \bigcap E_{2} = (3,H)$

 

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@GateOverflow04 Always solve such problem using Tree Method. It’s very easy. Conditional Probability is confusing at times.

The answer to your doubt : $E_{1} \bigcap E_{2} = (3,H)$ because only incase of Coin 3 we will get the reqd event when Toss comes Head and the other face is Tail. And it's $H$ because we need to find it's probability.

Answer 1

P(head) = 1(Doubly Headed coin) + 0 (Doubly tailed coin) + 1/2 (Normal coin)
P(Tail/Head) = P(one side is Tail and other is Head)/P(Head) = (1/2)/ (1+1/2+0)
=1/3

Answer 2

We toss a coin and it comes up as HEAD. So that "HEAD" is either of the "3 RED HEADS" in the tree above. The probability that the other face is tail is ?

Now, if the other face is tail, then the "HEAD" that was reported earlier must be from Coin 3. Therefore required probability is = 1/3

Answer 3

https://drive.google.com/file/d/0BxgoU2GSPq5pOGhOOXJmZ2YyYnM/view?usp=sharing

Answer 4

Using Bayes' theorem:-  

as head is on one side then the we need to find that tail is on other side .....for this it must be third coin....

P(C3/H)= P(C3) P(H/C3) / (P(H)  -> 1/3 * 1/2  / (1/3*1+1/3*0+1/3*1/2) ->1/3