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makhdoom ghaya
asked
in Quantitative Aptitude
Mar 18, 2016
recategorized
Apr 25, 2021
by Lakshman Patel RJIT

8,020 views
29 votes

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

- $\dfrac{1}{4}$
- $\dfrac{1}{3}$
- $\dfrac{1}{2}$
- $\dfrac{2}{3}$

@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.

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27 votes

Best answer

We are given $3$ coins out of which $1$ coin $(TT)$ is automatically removed from the total cases as we have already got a $H$.

Now when we toss any of two coins $(HH, HT),$ the cases for the front side are: $(H, H, H, T)$ and each of these are **equally likely**. It is given that the tossed side is $H$, so fourth option is also removed from the total cases.

Now as we have one coin with $HH$ and another coin with $HT$ and one $H$ side is present, the choices for another side are $(H, H, T).$

Now we want the $T$ side as favourable outcome out of total $3$ outcomes.

According to Bayes theorem or conditional probability

$p\left ( \text{another side is T} \right )=\dfrac{\text{# favourable outcomes}}{\text{#total outcomes}}$

$\therefore\; p\;=\dfrac{1}{3}$

So, answer is **B.**

**Reference link:** https://www.khanacademy.org/math/precalculus/prob-comb/dependent-events-precalc/v/bayes-theorem-visualized

50 votes

the answer is 1/3

we have to apply bayes theorem

A = H-H coin was chosen

B = T-T coin was chosen

C = T-H coin was chosen

E = Head comes up

now the probability of chosing the H-T coin when E has already happened

$P(\frac{C}{E})= \frac{P(C)*P(E/C)}{P(A)*P(E/A)+P(B)*P( E/B)+P(C)*P(E/C)}$

$P(A)=P(B)=P(C)=\frac{1}{3}$

$P(E/A)=1$

$P( E/C)=\frac{1}{2}$

$P( E/B)=0$

$P(\frac{C}{E})= \frac{\frac{1}{3}*\frac{1}{2}}{\frac{1}{3}*1+\frac{1}{3}*0+\frac{1}{3}*\frac{1}{2}}$

$P(\frac{C}{E})= \frac{1}{3}$

we have to apply bayes theorem

A = H-H coin was chosen

B = T-T coin was chosen

C = T-H coin was chosen

E = Head comes up

now the probability of chosing the H-T coin when E has already happened

$P(\frac{C}{E})= \frac{P(C)*P(E/C)}{P(A)*P(E/A)+P(B)*P( E/B)+P(C)*P(E/C)}$

$P(A)=P(B)=P(C)=\frac{1}{3}$

$P(E/A)=1$

$P( E/C)=\frac{1}{2}$

$P( E/B)=0$

$P(\frac{C}{E})= \frac{\frac{1}{3}*\frac{1}{2}}{\frac{1}{3}*1+\frac{1}{3}*0+\frac{1}{3}*\frac{1}{2}}$

$P(\frac{C}{E})= \frac{1}{3}$

9 votes

We are given that head comes on one side. So, there are only two possibilities- the other side is head ($\text{HH}$ coin) or the other side is tail ($\text{HT}$ coin). Favorable case is only $\text{HT}$, and hence answer is $\frac{1}{2}$.

Well, the above explanation is wrong because it assumes uniform probability distribution for the two cases $\text{HH}$ and $\text{HT}$. This is true initially as all 3 coins have equal probability of being chosen. But we are given that "head" comes in one face. $\text{HH}$ coin has two head faces, and hence has twice the chance of being chosen over $\text{HT}$ as we got a head face. i.e., for the two coins $\text{HH}$ and $\text{HT}$, the respective probabilities are $\frac{2}{3}$ and $\frac{1}{3}$ respectively given that one side is a head. Now, our required probability is simply $P(\{\text{HT}\}) = \frac{1}{3}$.

We can also apply Bayes' theorem but I guess solving intuitively is more fun :)

Well, the above explanation is wrong because it assumes uniform probability distribution for the two cases $\text{HH}$ and $\text{HT}$. This is true initially as all 3 coins have equal probability of being chosen. But we are given that "head" comes in one face. $\text{HH}$ coin has two head faces, and hence has twice the chance of being chosen over $\text{HT}$ as we got a head face. i.e., for the two coins $\text{HH}$ and $\text{HT}$, the respective probabilities are $\frac{2}{3}$ and $\frac{1}{3}$ respectively given that one side is a head. Now, our required probability is simply $P(\{\text{HT}\}) = \frac{1}{3}$.

We can also apply Bayes' theorem but I guess solving intuitively is more fun :)

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