Question

An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of 4 face values obtained, the probability that the minimum face value is not less than 2 and maximum face value is not more than 5 all the 4 times is \[\]
A. $ \dfrac{16}{81} $ \[\]
B. $ \dfrac{1}{81} $ \[\]
C. $ \dfrac{80}{81} $ \[\]
D. $ \dfrac{65}{81} $ \[\]

Answer 1

Hint: We find the probability for the roll of die once $ P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)} $ where $ n\left( A \right) $ the number of possible outcomes of a dice showing a number not less than 2 and not greater than 5 and $ n\left( S \right) $ is the sample size of all a dice roll. Since the dice roll one after another four times are independent events the required probability is $ {{\left( p\left( A \right) \right)}^{4}} $ .\[\]

Complete step by step answer:
We know from the definition of probability that if there is $ n\left( A \right) $ number of ways of event $ A $ occurring (or number of favourable outcomes) and $ n\left( S \right) $ is the size of the sample space (total number of outcomes) then the probability of the event $ A $ occurring is given by
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}\]
Independent events are events where the occurrence of one event does not affect the occurrence of other events. Mathematically, if $ A $ and $ B $ are two independent events then the probability of $ A $ and $ B $ happening simultaneously is given by,
\[P\left( A\bigcap B \right)=P\left( A \right).P\left( B \right)\]

We are given in the question that the unbiased die has its faces marked 1, 2, 3, 4, 5 and 6. So when it is rolled it will show one of its faces. Since there are only 6 numbers total number of outcomes is
 $ n\left( S \right)=6 $
Let us denote the event getting a face value not less than 2 and not greater than 5 as $ A $ in a single dice roll. We see that $ A $ can happen only when we get face values 2,3,4,5. So the number of possible outcome is
\[n\left( A \right)=4\]
 So the probability of event $ A $ happening is
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{4}{6}=\dfrac{2}{3}\]
We are given that the dice has been rolled 4 times. Since the roll of dice once does not affect the outcome of the successive dice roll, the 4 dice rolls are independent events and the probability that we a face value not less than 2 and not greater than 5 simultaneously is;
\[\begin{align}
  & P\left( A \right)\times P\left( A \right)\times P\left( A \right)\times P\left( A \right) \\
 & \Rightarrow \dfrac{2}{3}\times \dfrac{2}{3}\times \dfrac{2}{3}\times \dfrac{2}{3} \\
 & \Rightarrow {{\left( \dfrac{2}{3} \right)}^{4}}=\dfrac{16}{81} \\
\end{align}\]
So the required probability is $ \dfrac{16}{81} $ .The correct choice is A. \[\]

Note:
We must be careful of the difference between independent events where events happen together and mutually exclusive events where never happen together, for example getting a face value less than 2 and greater than 5 in a single dice roll. We can use the method of negation to get the required probability $ {{\left( 1-P\left( B \right) \right)}^{4}} $ where $ B $ is the event of a face value less than 2 and greater than 5 in a dice roll.