Question

For a biased die the probabilities for different faces to turn up are given below :

Face	$1$	$2$	$3$	$4$	$5$	$6$
Probability	$0.1$	$0.32$	$0.21$	$0.15$	$0.05$	$0.17$

The die is tossed and you are told that either face $1$ or $2$ has turned up. Then the probability that the face $1$ is

• A. $\frac{5}{21}$
• B. $\frac{5}{22}$
• C. $\frac{4}{21}$
• D. None of these

Answer 1

The correct option is A

$\frac{5}{21}$

Explanation for the correct option :

The probabilities of different faces of a biased die are given. Also, given that the die is tossed and either face $1$ or $2$ turned up.

We have to find the probability that the face is $1$.

Step-1 : Assumption

Let $A$be the event that the face $1$is turned up and $B$ be the event that the face $2$ is turned up.

Then, from the given table, we have $P\left(A\right)=0.1$ and $P\left(B\right)=0.32$.

Formula to be used : We know that for two mutually exclusive events $C$ and $D$, $P\left(C\cup D\right)=P\left(C\right)+P\left(D\right)$.

So, here, we must have

$$\begin{gathered} P\left(A\cup B\right)=P\left(A\right)+P\left(B\right) \\ =0.1+0.32 \\ =0.42 \end{gathered}$$

Step-2 : Finding the probability that the face is $1$ when it is known that either face $1$ or face $2$ turned up.

Clearly, here we have to find the conditional probability $P\left(A|A\cup B\right)$.

Formula to be used : We know that the conditional probability of an event $C$ when it is known that the event $D$ is already happened is given by

$P\left(C|D\right)=\frac{P(C\cap D)}{P\left(D\right)}$.

So, here the required probability would be

$$\begin{gathered} P\left(A|A\cup B\right)=\frac{P(A\cap (A\cup B\left)\right)}{P(A\cup B)} \\ =\frac{P\left(A\right)}{P\left(A\right)+P\left(B\right)}\left[\because (A\cap (A\cup B\left)\right)=A,P(A\cup B)=P\left(A\right)+P\left(B\right)\right] \\ =\frac{0.1}{0.1+0.32}\left[\because P\left(A\right)=0.1,P\left(B\right)=0.32\right] \\ =\frac{0.1}{0.42} \\ =\frac{5}{21} \end{gathered}$$

Hence, option (A) is the correct answer.