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.

Let us break down the solution step-by-step using the Binomial Probability Formula.

When a six-faced fair die is rolled, the probability of obtaining "ONE" on a single roll is:
$$p=\frac{1}{6}$$
Consequently, the probability of not obtaining "ONE" (failure) on a single roll is:
$$q=1-p=\frac{5}{6}$$

The die is rolled 5 times, so the number of trials is $n=5$. We want to find the probability of obtaining "ONE" at least four times. Let $X$ be the random variable representing the number of times "ONE" is obtained. We need to calculate:
$$P(X\ge 4)=P(X=4)+P(X=5)$$

Using the binomial probability formula, $P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}$, we calculate each term:

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

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

Adding the two probabilities:
$$P(X\ge 4)=\frac{25}{7776}+\frac{1}{7776}=\frac{26}{7776}\approx 0.0033436$$

To convert this probability to a percentage:
$$\text{Probability in \%}=0.0033436\times 100=0.33436\%$$
Rounding to two decimal places gives approximately 0.33%.