Question

A six-face unbiased die is thrown twice and the sum of the numbers appearing on the upper face is observed to be $7$ .the probability that the number $3$ has appears at least once, is
A. $\dfrac{1}{5}$
B. $\dfrac{1}{2}$
C. $\dfrac{1}{3}$
D. $\dfrac{1}{4}$

Answer 1

Hint: First, we need to analyze the given information carefully so that we are able to solve the problem. Here, we are given a probability consisting of an experiment. The experiment is throwing a six-faced die twice. We are asked to calculate the probability of the event of getting a number$3$ appearing at least once.
     We need to use the formula of the probability of an event in this question so that we can easily obtain the desired result.
Formula to be used:
      The formula to calculate the probability of an event is as follows.
The probability of an event (say A),$P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$

Complete step by step answer:
It is given that the six-faced die is thrown twice and this die contains$1,2,3,4,5,6$ on each face.
Also, it is given that the sum of the numbers that appeared on the upper facer will be$7$.
We shall first find that the sum of the numbers on the faces is$7$.
That is, we get\[\left( {1,6} \right),\left( {6,1} \right),\left( {2,5} \right),\left( {5,2} \right),\left( {3,4} \right),\left( {4,3} \right)\]
Therefore, there are six total outcomes.
We are asked to calculate the probability that the number$3$ appears at least once.
Here, the favorable outcome will be$\left( {3,4} \right)\left( {4,3} \right)$
Hence, there are two favorable outcomes for the required probability.
We shall apply the probability of an event.
The formula to calculate the probability of an event is as follows.
The probability of an event (say A),$P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$
Now, the probability that the number$3$ appears at least once$ = \dfrac{2}{6}$
                                                                                                               $ = \dfrac{1}{3}$
Hence the required probability is$\dfrac{1}{3}$

So, the correct answer is “Option C”.

Note: The probability of an event is nothing but the ratio of the number of favorable outcomes and the total number of outcomes. This is given by the formula$P\left( A \right) = \dfrac{{number{\text{ }}of{\text{ }}favorable{\text{ }}outcomes}}{{total{\text{ }}number{\text{ }}of{\text{ }}outcomes}}$.
Hence, we got the required probability of getting the number$3$ appearing at least once.