Question

A Ludo die is rolled, then the probability that the number of dots on its upper face is less than 4 is___.
A.$0$
B.$\dfrac{1}{2}$
C.$\dfrac{1}{3}$
D.$1$

Answer 1

Hint: We can find the probability of getting 1 dot, 2 dots and 3 dots separately. Then the probability that the number of dots on its upper face is less than 4 is given by the sum of the individual probability of getting numbers less than 4. We can take the sum of the probabilities and simplify it to get the required probability.

Complete step-by-step answer:
We know that a die has 6 faces and each face has a number of dots between 1 and 6.
So, the probability of getting 1 in the upper face is given by,
$P\left( 1 \right) = \dfrac{{{\text{no}}{\text{.}}\,{\text{of}}\,{\text{favouarble }}\,{\text{outcome}}}}{{{\text{no}}{\text{.}}\,{\text{of}}\,{\text{possible outcome}}}}$
$ \Rightarrow P\left( 1 \right) = \dfrac{1}{6}$
Similarly, we get, $P\left( 2 \right) = \dfrac{1}{6}$, $P\left( 3 \right) = \dfrac{1}{6}$
Probability of getting the number less than 4 is given by the sum of the probability of getting 1, 2 and 3 numbers of dots.
Let X be the number of dots on the upper face of the die. Then the probability of getting the number of dots in the upper face is less than 4 is given by,
$P\left( {X < 4} \right) = P\left( 1 \right) + P\left( 2 \right) + P\left( 3 \right)$
On substituting the values, we get
$P\left( {X < 4} \right) = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6}$
On taking the sum, we get
$P\left( {X < 4} \right) = \dfrac{3}{6}$
On dividing both numerator and denominator with 3, we get,
$P\left( {X < 4} \right) = \dfrac{1}{2}$
Therefore, the probability that the number of dots on its upper face is less than 4 is $\dfrac{1}{2}$.
So, the correct answer is option B.

Note: Alternate approach to the problem is given by,
We know that probability of an event is defined as the number of favourable outcomes divided by the total number of possible outcomes. For tossing a die, as there are 6 faces, the total number of possible outcomes is 6. For the event of getting a number less than 4, the possible outcomes are getting 1,2 and 3. So the number of favourable outcomes is 3. So the required probability is given by,
$P = \dfrac{{{\text{no}}{\text{.}}\,{\text{of}}\,{\text{favourable }}\,{\text{outcome}}}}{{{\text{no}}{\text{.}}\,{\text{of}}\,{\text{possible outcome}}}}$
$ \Rightarrow P = \dfrac{3}{6} = \dfrac{1}{2}$
Therefore, the probability that the number of dots on its upper face is less than 4 is $\dfrac{1}{2}$.