Question

In a bag there are 44 identical cards with figure of circle or square on them. There are 24 circles, of which 9 are blue and rest are green and 20 squares of which 11 are blue and rest are green. One card is drawn from the bag at random. Find the probability that it has the figure of
(i) square (ii) green colour (iii) blue circle and (iv) green square. [CBSE 2015]

Answer 1

There are 44 identical cards with figure of circle or square in the bag.

∴ Total number of outcomes = 44

(i) There are 20 squares on the cards.

So, the favourable number of outcomes is 20.

∴ P(card drawn has the figure of square) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{20}{44}=\frac{5}{11}$

(ii) Number of cards with green circle = 24 − 9 = 15

Number of cards with green square = 20 − 11 = 9

∴ Number of cards with green colour figure = 15 + 9 = 24

So, the favourable number of outcomes is 24.

∴ P(card drawn has the figure of green colour) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{24}{44}=\frac{6}{11}$

(iii) There are 9 cards with circle of blue colour.

So, the favourable number of outcomes is 9.

∴ P(card drawn has the figure of blue circle) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{9}{44}$

(iv) Number of cards with green square = 20 − 11 = 9

So, the favourable number of outcomes is 9.

∴ P(card drawn has the figure of green square) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{9}{44}$