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

Answer :

we know Probability P ( E ) = $\frac{Total number of desired events n ( E )}{Total number of events n ( S )}$

Here Total number of events ( Total games )  n (  S ) =   8 + 10  =  18

i ) One piece is lost at random. Find the probability that it is a triangle .

So,

Total number of triangles =  8  , So n ( E ) = 8 

And
Probability  = $\frac{8}{18} = \frac{\mathbf{4}}{\mathbf{9}} \mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)}$

ii ) One piece is lost at random. Find the probability that it is a square .

So,

Total number of squares  = 10  , So n ( E ) = 10

And
Probability  = $\frac{10}{18} = \frac{\mathbf{5}}{\mathbf{9}} \mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)}$

iii ) One piece is lost at random. Find the probability that it is a square of blue of colour .

So,

Total number of square of blue of colour  =  6  , So n ( E ) = 6

And
Probability  = $\frac{6}{18} = \frac{\mathbf{1}}{\mathbf{3}} \mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)}$

iv ) One piece is lost at random. Find the probability that it is a triangle of red colour .

So,

Total number of triangle of red colour  =  5  , So n ( E ) = 5

And
Probability  = $\frac{\mathbf{5}}{\mathbf{18}} \mathbf{(}\mathbf{ }\mathit{A}\mathit{n}\mathit{s}\mathbf{ }\mathbf{)}$