Question

If $x$ denotes the number of sixes in four consecutive throws of a dice, then $P\left( {x = 4} \right)$ is
1. $\dfrac{1}{{1296}}$
2. $\dfrac{4}{6}$
3. $1$
4. $\dfrac{{1295}}{{1296}}$

Answer 1

Hint: In this question, we are given that the number of sixes in four consecutive throws of dice is $x$ and we have to find the probability that in four consecutive throws of a dice number of sixes is four. The first step is to calculate the probability of $6$will occur when we throw a die ($p$). Then using $p + q = 1$ find the probability of $6$ will not occur. In last apply binomial distribution to find the probability $P\left( {x = 4} \right)$.

Formula Used:
Probability formula – The probability of an event occurring is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
\[Probability = \dfrac{{Number{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
Binomial distribution –
$P\left( {x = r} \right) = {}^n{C_r}{p^r}{q^{n - r}}$
Where, $p,q$are the number of successes and failures respectively.
Combination formula –
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Complete step by step Solution:
Given that,
The total number of sixes in four consecutive throws of dice is $x$,
Numbers in a die are $1,2,3,4,5,6$
Number of possible outcomes$ = 6$
Total number of chances that $6$ will occur$ = 1$
Number of favorable outcomes $ = 1$
Using the probability formula,
\[Probability = \dfrac{{Number{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
Probability that $6$will occurs is $p = \dfrac{1}{6}$
And probability that $6$will not occur will be $q = 1 - p = \dfrac{5}{6}$
Also, the number of sixes in four consecutive throws of a dice.
It implies that, $n = 4$
Using the binomial distribution,
$P\left( {x = r} \right) = {}^n{C_r}{p^r}{q^{n - r}}$
Therefore, $P\left( {x = 4} \right) = {}^4{C_4}{\left( {\dfrac{1}{6}} \right)^4}{\left( {\dfrac{5}{6}} \right)^{4 - 4}}$
Using combination formula, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$P\left( {x = 4} \right) = \dfrac{{4!}}{{4!\left( {4 - 4} \right)!}}\left( {\dfrac{1}{{1296}}} \right)\left( 1 \right)$
$P\left( {x = 4} \right) = \dfrac{1}{{1296}}$

Hence, the correct option is 1.

Note: The key concept involved in solving this problem is a good knowledge of Binomial distribution. Students must remember that the binomial distribution with parameters $n$ and $p$ is the discrete probability distribution of the number of successes in a series of $n$ independent experiments, each of which asks a yes-no question and has its own Boolean-valued outcome i.e., success or failure. The binomial distribution can be used to calculate the likelihood of obtaining a certain number of successes, such as successful football goals, from a fixed number of trials.