Question

A child’s game has 8 triangles of which 3 are blue and the rest are red and 10 squares of which 6 are blue and the rest are red. One piece is lost at random. Find the probability that it is a
(i) triangle
(ii) square
(iii) square of blue color
(iv) triangle of red color

Answer 1

Hint: In order to solve the problems based on probability, the first thing one can do is to find the Total number of possible outcomes. Here, in this case, the total number of possible outcomes refers to the total number of figures or shapes available. Now, we need to find the probability of an event using the total number of outcomes favorable to the occurrence of the particular event. We need to remember the classical formula to calculate the probability of an event which is given by –
$\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{n}\left( \text{A} \right)}{\text{n}\left( S \right)}....................(i)$
where A is the event that is to occur
P (A) is the probability of occurrence of event A
n (A) is the total number of outcomes favorable to the occurrence of event A
n (S) is the total number of equally likely events or the total number of possible outcomes

Complete step-by-step solution:
Here, it is given that the child’s game has 8 triangles of which 3 are blue and the rest are red, and 10 squares of which 6 are blue and the rest are red. So,
Total number of equally likely events, $\text{n}\left( \text{Figure} \right)=10+8=18$
Also, it is given that one piece is lost at random.
Now,
(i) We have to find the probability that the lost piece is a triangle.
Since, there are total of 8 triangles, so,
Total number of outcomes favorable to the occurrence of the event, $\text{n}\left( \text{triangle} \right)=8$
Now, the probability that the lost piece is a triangle is given by –
\[\text{P}\left( \text{triangle} \right)\text{=}\dfrac{\text{n}\left( \text{triangle} \right)}{\text{n}\left( \text{Figure} \right)}....................(ii)\]
Now, replacing the value of $\text{n}\left( \text{Figure} \right)$ and $\text{n}\left( \text{Triangle} \right)$ with their respective values in equation (ii), we get,
\[\Rightarrow \text{P}\left( \text{triangle} \right)\text{=}\dfrac{8}{18}\]
\[\Rightarrow \text{P}\left( \text{triangle} \right)\text{=}\dfrac{4}{9}\]
Hence, the probability that the lost piece is a triangle is $\dfrac{4}{9}$.
Again,
(ii) We have to find the probability that the lost piece is a square.
Since, there are total of 10 squares, so,
Total number of outcomes favorable to the occurrence of the event, \[\text{n}\left( \text{square} \right)=10\]
Now, the probability that the lost piece is a square is given by –
\[\text{P}\left( \text{square} \right)\text{=}\dfrac{\text{n}\left( \text{square} \right)}{\text{n}\left( \text{Figure} \right)}....................(iii)\]
Now, replacing the value of $\text{n}\left( \text{Figure} \right)$ and $\text{n}\left( \text{Square} \right)$ with their respective values in equation (iii), we get,
\[\Rightarrow \text{P}\left( \text{square} \right)\text{=}\dfrac{10}{18}\]
\[\Rightarrow \text{P}\left( \text{square} \right)\text{=}\dfrac{5}{9}\]
Hence, the probability that the lost piece is a square is $\dfrac{5}{9}$.
Also,
(iii) We have to find the probability that the lost piece is a square of blue color.
Since, out of 10 squares, 6 are blue, so,
Total number of outcomes favorable to the occurrence of the event, \[\text{n}\left( \text{square of blue color} \right)=6\]
Now, the probability that the lost piece is a square of blue color is given by –
\[\text{P}\left( \text{square of blue color} \right)\text{=}\dfrac{\text{n}\left( \text{square of blue color} \right)}{\text{n}\left( \text{Figure} \right)}....................(iv)\]
Now, replacing the value of $\text{n}\left( \text{Figure} \right)$ and $\text{n}\left( \text{Square of blue color} \right)$ with their respective values in equation (iv), we get,
\[\Rightarrow \text{P}\left( \text{Square of blue color} \right)\text{=}\dfrac{6}{18}\]
\[\Rightarrow \text{P}\left( \text{Square of blue color} \right)\text{=}\dfrac{1}{3}\]
Hence, the probability that the square of blue colour is $\dfrac{1}{3}$.
Finally,
(iv) We have to find the probability that the lost piece is a triangle of red color.
Since, out of 8 triangles, 3 are blue and rest are red, i.e., 5 are red, so,
Total number of outcomes favorable to the occurrence of the event, $\text{n}\left( \text{triangle of red color} \right)=5$
Now, the probability that the lost piece is a triangle of red color is given by –
\[\text{P}\left( \text{triangle of red color} \right)\text{=}\dfrac{\text{n}\left( \text{triangle of red color} \right)}{\text{n}\left( \text{Figure} \right)}....................(v)\]
Now, replacing the value of $\text{n}\left( \text{Figure} \right)$ and $\text{n}\left( \text{Triangle of red color} \right)$ with their respective values in equation (ii), we get,
\[\Rightarrow \text{P}\left( \text{triangle of red color} \right)\text{=}\dfrac{5}{18}\]
\[\Rightarrow \text{P}\left( \text{triangle of red color} \right)\text{=}\dfrac{5}{18}\]
Hence, the probability that the lost piece is a triangle of red color is $\dfrac{5}{18}$.

Note: Students often make mistake in differentiating between the number of particular entities. Students carefully need to find the total number of possible outcomes by adding only the main category of entities. Generally, students add all the numbers seen in question, i.e., $8+3+10+6=27$, which is completely wrong. Besides, students need carefully understand that out of 8 triangles, it is said that 3 are blue, and the rest are red. This implies are that we need to find the number of triangles which are red in color and it is obtained by subtracting the total number triangles with the number of triangles which are blue in color, i.e., $8-3=5$.