A six-faced fair dice is rolled five times. The probability (in %) of obtaining “ONE” at least four times is

The correct option is 0.33.

Step-by-Step Explanation:

Let $X$ be the random variable representing the number of times "ONE" is obtained when a six-faced fair die is rolled 5 times. Since each roll is independent and has the same probability of success, $X$ follows a binomial distribution:
$X~\text{Binomial}(n,p)$

The parameter values are:
Number of trials, $n=5$
Probability of obtaining "ONE" in a single roll (success), $p=\frac{1}{6}$
Probability of not obtaining "ONE" in a single roll (failure), $q=1-p=\frac{5}{6}$

The probability of getting exactly $k$ successes in $n$ trials is given by the binomial probability formula:
$P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}$
where $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$ is the binomial coefficient.

We want to find the probability of obtaining "ONE" at least four times, which is $P(X\ge 4)$:
$P(X\ge 4)=P(X=4)+P(X=5)$

Let's calculate each term:
1) For $k=4$:
$P(X=4)=\left(\genfrac{}{}{0ex}{}{5}{4}\right){\left(\frac{1}{6}\right)}^4{\left(\frac{5}{6}\right)}^1$
$P(X=4)=5\times \frac{1}{1296}\times \frac{5}{6}=\frac{25}{7776}$

2) For $k=5$:
$P(X=5)=\left(\genfrac{}{}{0ex}{}{5}{5}\right){\left(\frac{1}{6}\right)}^5{\left(\frac{5}{6}\right)}^0$
$P(X=5)=1\times \frac{1}{7776}\times 1=\frac{1}{7776}$

Now, we add the two probabilities:
$P(X\ge 4)=\frac{25}{7776}+\frac{1}{7776}=\frac{26}{7776}=\frac{13}{3888}$
$P(X\ge 4)\approx 0.0033436$

Converting the probability to percentage:
$\text{Percentage}=0.0033436\times 100\%\approx 0.33\%$