Question

If a fair die is rolled 4 times, then what is the probability that there are at least two sixes?
A $\dfrac{19}{144}$
B $\dfrac{25}{216}$
C $\dfrac{125}{216}$
D $\dfrac{175}{216}$

Answer 1

In this question we have been given a situation for which we need to calculate the probability by binomial distribution. So, for that firstly we will be calculating the probability of getting 6 and after that the probability of not getting 6 by the formula:
Probability = favorable outcomes ÷ total number of outcomes
So according to the question the number of trials would be equal to 4. In the next step we will be using the formula of binomial distribution $P(x=r)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$. So, as we need to calculate the probability for at least two sixes, we will be calculating the probability for r equal to 2,3 and 4 and add them which will be our desired solution.

Complete step by step answer:
We have been provided with a condition in this question, so first we need to write the sample space of a dice
Sample space of a dice= (1,2,3,4,5,6)
Now we will be calculating the probability of getting 6 and the probability of not getting 6 by using the formula: Probability = favorable outcomes $\div$ total number of outcomes
Using the above formula, we get
Probability of getting 6, i.e p = $\dfrac{1}{6}$
Probability of not getting 6, i.e q = $\dfrac{5}{6}$
Now according to the question, the die was thrown 4 times so the number of trials is equal to 4.
Number of trials n=4
We will use the formula of binomial distribution, $P(x=r)={}^{n}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$ .
Now as in this question we need to calculate the probability for at least 2 sixes so it means we can take the value of r=2,3,4 respectively.
For r=2, substituting values of n, r, p and q, we have ${}^{4}{{C}_{2}}{{\left( \dfrac{1}{6} \right)}^{2}}{{\left( \dfrac{5}{6} \right)}^{2}}$ .
Now we will be simplifying it and finding its value which comes out to be $\dfrac{150}{1296}$ .
Now we will be calculating for r=3, as ${}^{4}{{C}_{3}}{{\left( \dfrac{1}{6} \right)}^{3}}{{\left( \dfrac{5}{6} \right)}^{1}}$ .
Now we will be simplifying it and finding its value which comes out to be $\dfrac{20}{1296}$ .
Now we will be calculating for r=4, as ${}^{4}{{C}_{4}}{{\left( \dfrac{1}{6} \right)}^{4}}{{\left( \dfrac{5}{6} \right)}^{0}}$ .
Now we will be simplifying it and finding its value which comes out to be $\dfrac{1}{1296}$ .
Now after finding all the probabilities we will be adding them as we know for the probability of at least two sixes all the above cases would be considered
$\dfrac{150}{1296}+\dfrac{20}{1296}+\dfrac{1}{1296}=\dfrac{171}{1296}$
Now we will simplify the answer a little bit and then the final answer would be $\dfrac{19}{144}$ .
So, the answer comes out to be $\dfrac{19}{144}$ which is your option number (a) .

So, the correct answer is “Option A”.

Note: We can solve the above question by another method in which we will be calculating the probability of at most two sixes by the same method mentioned above and after that subtracting it from 1. So, this could be another method of solving the same question.