Question

In throwing 3 dice, the probability that at least 2 of the three numbers obtained are same is
A) ${\displaystyle \frac{1}{2}}$
B) ${\displaystyle \frac{1}{3}}$
C) ${\displaystyle \frac{4}{9}}$
D) None

Answer 1

Hint: For finding the probability that at least 2 of the three numbers obtained are the same will be the summation of the probability of exactly 2 of the three numbers are same and the probability of all the three numbers obtained are same. For that find, the probability of no numbers is the same on any dice and subtract from 1 to get the desired result.

Complete step-by-step solution:
Given: - 3 dice are thrown.
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. how likely they are to happen, using it.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favorable outcomes and the total number of outcomes.
$P\left(E\right)={\displaystyle \frac{n\left(E\right)}{n\left(S\right)}}$
We know that the total number of possible outcomes when three dice are thrown is
$\Rightarrow n\left(S\right)=6\times 6\times 6$
Multiply the terms,
$\Rightarrow n\left(S\right)=216$...............….. (1)
For finding the probability that at least 2 of the three numbers obtained are the same will be the summation of the probability that exactly 2 of the three numbers are the same and the probability of all the three numbers obtained are the same.
So, find the probability of no numbers is the same on any dice and subtract from 1.
The favorable outcomes for no numbers are the same is,
$\Rightarrow n\left(E\right)=6\times 5\times 4$
Multiply the terms,
$\Rightarrow n\left(E\right)=120$.................….. (2)
Now, substitute the values from equation (1) and (2) to find the probability of no numbers is the same is,
$\Rightarrow P\left(E\right)={\displaystyle \frac{120}{216}}$
Cancel out the common factors,
$\Rightarrow P\left(E\right)={\displaystyle \frac{5}{9}}$
Now subtract the probability from 1 to get the probability that at least 2 of the three numbers obtained are the same.
$\Rightarrow P\left(\overline{E}\right)=1-P\left(E\right)$
Substitute the value of $P\left(E\right)$,
$\Rightarrow P\left(\overline{E}\right)=1-{\displaystyle \frac{5}{9}}$
Take LCM on the right side,
$\Rightarrow P\left(\overline{E}\right)={\displaystyle \frac{9-5}{9}}$
Subtract the values in the numerator,
$\therefore P\left(\overline{E}\right)={\displaystyle \frac{4}{9}}$
Hence, the probability that at least 2 of the three numbers obtained are the same is ${\displaystyle \frac{4}{9}}$.

Option C is the correct answer.

Note: We must calculate the number of favorable and possible outcomes in each case to calculate the probability of each of the given events. We should also be careful that we don’t count the same event repeatedly or we miss some event.