Question

 A person throws two dice, one the common cube, and the other a regular tetrahedron, the number on the lowest face being taken in the case of the tetrahedron; what is the chance that the number being thrown is not less than 5?

Answer 1

Hint: A regular tetrahedron has 4 faces and a regular cube has 6 faces. This means, the numbers on faces of tetrahedron will be from 1 to 4, whereas on the common cube dice will be from 1 to 6. To solve this question, we will calculate the probability of getting the numbers with sum less than 5 and then subtract it from 1, as the sum of probability of having an event true and probability of having an event false is one.

Complete step by step answer:
Let S be the event of throwing both dice and getting a number at random.
S = $\left\{ \begin{align}
  & \left( 1,1 \right),\left( 1,2 \right),\left( 1,3 \right),\left( 1,4 \right) \\
 & \left( 2,1 \right),\left( 2,2 \right),\left( 2,3 \right),\left( 2,4 \right) \\
 & \left( 3,1 \right),\left( 3,2 \right),\left( 3,3 \right),\left( 3,4 \right) \\
 & \left( 4,1 \right),\left( 4,2 \right),\left( 4,3 \right),\left( 4,4 \right) \\
 & \left( 5,1 \right),\left( 5,2 \right),\left( 5,3 \right),\left( 5,4 \right) \\
 & \left( 6,1 \right),\left( 6,2 \right),\left( 6,3 \right),\left( 6,4 \right) \\
\end{align} \right\}$
Let n(S) be the number of elements in event set S.
Therefore, n(S) = 24
Let A be the event of getting the sum less than 5.
A = $\left\{ \begin{align}
  & \left( 1,1 \right),\left( 1,2 \right),\left( 1,3 \right) \\
 & \left( 2,1 \right),\left( 2,2 \right) \\
 & \left( 3,1 \right) \\
\end{align} \right\}$
Let n(A) be the number of elements in event set A.
Therefore, n(A) = 6
The probability of getting the sum of drawn numbers less than 5 will be given P(A) = $\dfrac{\text{n}\left( \text{A} \right)}{\text{n}\left( \text{S} \right)}$, where P(A) is the probability of the event A.
Therefore, P(A) =$\dfrac{6}{24}=\dfrac{1}{4}$.
The probability of an event not happening is given by P(A’) = 1 – P(A), where A’ is the event of event A not happening.
But, P(A) = $\dfrac{1}{4}$, therefore, P(A’) = 1 – $\dfrac{1}{4}$= $\dfrac{3}{4}$.
Therefore, the probability of not getting the sum less than 5 is $\dfrac{3}{4}$.

Note: Students with good practice in sets and probability can omit writing the complete set and directly write the number of elements in the set. It is advisable to be vigilant about the way the dice are thrown. In this case, they are thrown simultaneously and hence, the sample set has 24 elements. If the way the dice are thrown changes, the number of elements in the sample set may change.