Question

A game has 8 triangles of which 6 are blue and rest are green, 12 rectangles of which 3 are green and rest are blue, and 10 rhombuses of which 3 are blue and rest are green. One piece is lost at random. Find the probability that it is (i) a rectangle (ii) a triangle of green colour (iii) a rhombus of blue colour

Answer 1

Explanation:

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Answer 2

Question:

A game has 8 triangles of which 6 are blue and rest are green, 12 rectangles of which 3 are green and rest are blue, and 10 rhombuses of which 3 are blue and rest are green. One piece is lost at random. Find the probability that it is (i) a rectangle (ii) a triangle of green colour (iii) a rhombus of blue colour.

Answer:

(i) The probability of losing a rectangle is $\dfrac{2}{5}$.

(ii) The probability of losing a green triangle is $\dfrac{1}{15}$.

(iii) The probability of losing a blue rhombus is $\dfrac{1}{10}$

Given:

The number of triangles is 8.

Out of 8 triangles, 6 are blue and rests are green.

The number of rectangle is 12.

Out of 12 rectangles, 3 are green and rests are blue.

The number of rhombuses is 10.

Out of 10 rhombuses, 3 are blue and rests are green.

Among the triangle, rectangles and rhombus, 1 piece lost at random.

To find:

(i) The probability of losing a rectangle.

(ii) The probability of losing a green triangle.

(iii) The probability of losing a blue rhombus.

Explanation:

The number of triangles is 8.

The number of rectangle is 12.

The number of rhombuses is 10.

Hence, we get the total number of outcome as _

(8+12+10)

= 30.

∴ In each case, total outcome = 30.

We know that,

$Probability =\dfrac{Total number of outcome}{ The Number of favorable outcome}$

(i)

The number of rectangle is 12.

∴ The number of favorable outcome is 12.

∴ P.E.(getting a rectangle) = $\frac{12}{30}$

                                          = $\frac{2}{5}$ [∵ Dividing 12 and 30 by 6.]

∴ The probability of losing a rectangle is $\frac{2}{5}$ .

(ii)

The number of triangles is 8.

The number of blue triangles is 6.

∴The number of green triangles is (8 - 6)

                                                         = 2.

∴ The number of favorable outcome is 2.

∴ P.E.(getting a green triangle) = $\frac{2}{30}$

                                                   = $\frac{1}{15}$ [∵ Dividing 2 and 30 by 2.]

∴ The probability of losing a green triangle is $\frac{1}{15}$

(iii)

The number of rhombuses is 10.

The number of blue rhombuses is 3.

∴ The number of favorable outcome is 3.

∴ P.E.(getting a blue rhombus) = $\frac{3}{30}$

                                                  = $\frac{1}{10}$ [∵ Dividing 3 and 30 by 3.]

∴ The probability of losing a blue rhombus is $\frac{1}{10}$