Question

Four fair dice D1, D2, D3, D4 each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously. The probability that D4 shows a number appearing on one of D1, D2 and D3 is

1. $\dfrac{{91}}{{216}}$
2. $\dfrac{{108}}{{216}}$
3. $\dfrac{{125}}{{216}}$
4. $\dfrac{{127}}{{216}}$

Answer 1

Hint : Here, we are given 4 dice that roll simultaneously. Each dice has 6 numbers in it (1 to 6) and so the total number of outcomes when n dice are rolled is ${6^n}$ . We need to find the required probability is = 1 – P (Dice D4 shows none of dice D1, D2 and D3).

Complete step-by-step answer :
Given that,
There are four fair dice rolled simultaneously.
We know that,
Probability of occurs if event E is
$P(E) = \dfrac{{No.{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Total{\text{ }}no.{\text{ }}of{\text{ }}outcomes}}$
Also, the sum of the probabilities is equal to 1. i.e. n(S)=1
Total number of outcomes when n dice are rolled is ${6^n}$ .
Let S be the sample space.
Assume that the number shown on dice D4 is ‘a’.
Let E’ be the event that none of the dices D1, D2 and D3 show number ‘a’ and E be the event that one of the dice (at least one) D1, D2 and D3 show number ‘a’.
This means, S = E + E’
Since, in the event ${E'}$ the other dice can show any number other than the number ‘a’.
Number of favourable cases of ${E'}$ is
$n({E'}) = {5^3}$
Because, each dice has 5 favourable possibilities & there are 3 dice with which comparison is done.
Also,
Total number of outcomes when 3 dice are rolled is
$n(S) = {6^3}$
Probability of the event ${E'}$ is
$P({E'}) = \dfrac{{{5^3}}}{{{6^3}}}$
We have,
$P(E) + P({E'}) = 1$
$\Rightarrow P(E) = 1 - P({E'})$
$\Rightarrow P(E) = 1 - \dfrac{{{5^3}}}{{{6^3}}}$
$\Rightarrow P(E) = 1 - {(\dfrac{5}{6})^3}$
$\Rightarrow P(E) = 1 - \dfrac{{125}}{{216}}$
$\Rightarrow P(E) = \dfrac{{216 - 125}}{{216}}$
$\Rightarrow P(E) = \dfrac{{91}}{{216}}$

Another Method:
Required probability is
= 1 – P(Dice D4 shows none of dice D1, D2 and D3)
Since, a dice has 6 numbers.
Thus, D4 can show any of the six numbers.
According to the given information,
If D4 shows a number ‘a’ then dice D1, D2 and D3 show any other number except number ‘a’.
Thus, the required probability is
$= 1 - {}^6{C_1} \times \dfrac{1}{6} \times {(\dfrac{5}{6})^3}$
$= 1 - \dfrac{{6 \times {5^3}}}{{6 \times {6^3}}}$
$= 1 - \dfrac{{{5^3}}}{{{6^3}}}$
Hence, the probability that D4 shows a number appearing on one of D1, D2 and D3 is $\dfrac{{91}}{{216}}$ .
So, the correct answer is “Option 1”.

Note : Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination.