Question

For a biased dice, the probabilities for the different faces to turn up are

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

The dice is tossed and it is told that either the face 1 or face 2 has shown up, then the probability that it is face 1 is
A) $\dfrac{16}{21}$
B) $\dfrac{1}{10}$
C) $\dfrac{5}{10}$
D) $\dfrac{5}{21}$

Answer 1

Hint: We will use the formula $P\left( \dfrac{1}{E} \right)=\dfrac{1\cap E}{E}$, where 1 is the probability of face 1 which is 0.1 and ‘E’ is the total probability. We will find first $P\left( E \right)$ by adding $P\left( 1 \right)+P\left( 2 \right)$ and put the value accordingly to find the probability that it is face one.

Complete step-by-step answer:
It is given in the question that for a biased die, the probability for the different faces to turn up as –

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

It is also said that when die is tossed either face 1 or face 2 are turned up, then we have to find the probability that it is face one.
Let us assume that the probability of face one being $P\left( A \right)$, also assume that the probability of face 2 be $P\left( B \right)$ and $P\left( S \right)$ be the probability of face one and face two together.
We are given that Probability of first face $P\left( A \right)=0.1$ and the probability of second face, $P\left( B \right)=0.32$ then the total probability of face one and face two together will be, $P\left( S \right)=P\left( A \right)+P\left( B \right)$ that is, $P\left( S \right)=0.1+0.32=0.42$.
Now, we will find the probability of face on $P\left( A \right)$ using the formula $P\left( \dfrac{A}{S} \right)=\dfrac{P\left( A\cap S \right)}{P\left( S \right)}$. We know that $P\left( A\cap S \right)=0.1$ and $P\left( S \right)=0.42$ thus on putting these values in the above formulae, we get
$P\left( \dfrac{A}{S} \right)=\dfrac{0.1}{0.42}=\dfrac{10}{42}$, further solving, we get
$P\left( \dfrac{A}{S} \right)=\dfrac{5}{21}$,
Thus option d) is the correct answer.

Note: Students may take $P\left( S \right)$ as the total sum of all the faces which is always 1, and this is not at all correct because it is clearly mentioned in the question that face one or face two will appear in that particular throw. Thus, it is recommended to read and understand the question carefully before solving it.