Question

A purse contains 2 six-sided dice. One is a normal fair die, while the other has two 1′s, two 3′s and two 5′, a die is picked up and rolled. Because of some secret magnetic attraction of the unfair die, there is \[75\% \] chance of picking the unfair die and a \[25\% \] chance of picking a fair die. The die is rolled and shows up the face 3. The probability that a fair die was picked up is

Answer 1

Hint:In this question the given purse contains 2 six-sided dice. One is a normal fair die, the other has two 1′s, two 3′s and two 5′s and also they give the percentage chance of picking a fair and unfair die. First we have to find the required probability of chance picking of fair or unfair.

Complete step-by-step answer:
Here, it is given that there is \[75\% \] chance of picking the unfair die and a \[25\% \] chance of picking a fair die
Let N be the normal die picked, and M be the magnetic die picked and also let A be die shows up 3.
According to the question,
\[P\left( N \right) = \dfrac{1}{4}\]
\[P\left( M \right) = \dfrac{3}{4}\]
There is \[75\% \] chance of picking the unfair die and a \[25\% \] chance of picking a fair die,
for fair die, probability of choosing 3
\[P\left( {\dfrac{A}{N}} \right)\]= \[\dfrac{1}{6}\]
And again
For unfair die, probability of choosing 3
\[P\left( {\dfrac{A}{M}} \right)\] \[ = \dfrac{2}{6} = \dfrac{1}{3}\]
We know that the probability that Events $A$ and $B$ both occur is the probability of the intersection of $A$ and $B$. The probability of the intersection of Events $A$ and $B$ is denoted by \[P\left( {A\; \cap \;B} \right)\]. If Events A and B are mutually exclusive, \[P\left( {A\; \cap \;B} \right) = 0\].
We can write \[P\left( {A \cap N} \right)\] and \[P\left( {A \cap M} \right)\]
\[P\left( A \right) = P\left( {A \cap N} \right) + P\left( {A \cap M} \right)\]
\[ \Rightarrow P\left( A \right) = P\left( N \right) \times P\left( {\dfrac{A}{N}} \right) + P\left( M \right) \times P\left( {\dfrac{A}{M}} \right)\]
\[ \Rightarrow P\left( A \right) = \dfrac{1}{4} \times \dfrac{1}{6} + \dfrac{3}{4} \times \dfrac{1}{3}\]
\[ \Rightarrow P\left( A \right) = \dfrac{1}{{24}} + \dfrac{1}{4} = \dfrac{7}{{24}}\]
The probability that a fair die was picked up is
​ \[P\left( {\dfrac{N}{A}} \right) = \dfrac{{P\left( {N \cap A} \right)}}{{P\left( A \right)}} = \dfrac{{\dfrac{1}{4} \times \dfrac{1}{6}}}{{\dfrac{7}{{24}}}}\]
Solving and we get
\[P\left( {\dfrac{N}{A}} \right) = \dfrac{1}{7}\]
Hence, the required probability that a fair die was picked up is \[\dfrac{1}{7}\].

Note:The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty. The higher the probability of an event, the more likely it is that the event will occur.