GO Classes 2024 | Weekly Quiz 8 | Conditional Probability | Question: 10

Answer 1

I chosen two $dice$ at random (both are equally likely)
So, probability of choosing the dice will be equal, $P(D_{1}) = P(D_{2}) = 0.5 = \frac{1}{2}$

$D_{1}$ has $3$ red & $3$ black faces; $D_{2}$ has $1$ red & $5$ black faces.

So, $P(R|D_{1}) = \frac{3}{6} = \frac{1}{2},  P(B|D_{1}) = \frac{1}{2}$ & $P(R|D_{2}) = \frac{1}{6}, P(B|D_{2}) = \frac{5}{6}$

So, probability that I chose the die with $3$ red faces, given that first roll came up “$red$”:

$P(D_{1}|R) = \frac{P(D_{1}\cap R)}{P(R)} = \frac{P(R|D_{1})\times P(D_{1})}{P(R|D_{1})\times P(D_{1}) + P(R|D_{2})\times P(D_{2})} = \frac{(\frac{1}{2}) \times (\frac{1}{2})}{(\frac{1}{2}) \times (\frac{1}{2}) + (\frac{1}{6}) \times (\frac{1}{2})} = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{6}} = \frac{3}{4}$

${\color{Green} Ans: B. 3/4}$  

Comments

Full marks for your efforts dude!

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@chokostar Thank you; It means a lot 😊

Answer 2

We use Bayes' theorem. Let $C_1$ be the event I chose the first die, and $C_2$ be the event I chose the second die. By assumption
$$
P\left(C_1\right)=P\left(C_2\right)=\frac{1}{2}
$$
Let $A$ be the event I roll a red on the first roll. We're told how many red faces each die has, so we have
$$
P\left(A \mid C_1\right)=\frac{3}{6}=\frac{1}{2} \text { and } P\left(A \mid C_2\right)=\frac{1}{6}
$$
By Bayes' Theorem,
$$
P\left(C_1 \mid A\right)=\frac{P\left(A \mid C_1\right) P\left(C_1\right)}{P\left(A \mid C_1\right) P\left(C_1\right)+P\left(A \mid C_2\right) P\left(C_2\right)}=\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{2}\right)}=\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{6}}=\frac{3}{4}.
$$