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

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Solution

No. of cards in the bag $= 44$
No. of cards with circles $= 24$
No. of cards with blue circles $= 9$
No. of cards with green circles $= 24 - 9 = 15$
No. of cards with squares $= 20$
No. of cards with blue squares $= 11$
No. of cards with green squares $= 9$

Solution(i):
Let $E$ be the event of drawing a square card from the bag

We know that, Probability $P (E)$ $= \frac{(No.of favorable outcomes)}{(Total no.of possible outcomes)}$ $= \frac{20}{44}$ $= \frac{5}{11}$

Therefore, the Probability that the square card is drawn $= \frac{5}{11}$

Solution(ii):
Let $E$ be the event of drawing a green card from the bag

No. of green cards $= 15 + 9 = 24$

We know that, Probability $P (E)$ $= \frac{(No.of favorable outcomes)}{(Total no.of possible outcomes)}$ $= \frac{24}{44}$ $= \frac{6}{11}$

Therefore, the Probability that the green card is drawn $= \frac{6}{11}$

Solution(iii):
Let $E$ be the event of drawing a blue circle from the bag

We know that, Probability $P (E)$ $= \frac{(No.of favorable outcomes)}{(Total no.of possible outcomes)}$ $= \frac{9}{44}$

Therefore, the Probability that the blue circle is drawn $= \frac{9}{44}$

Solution(iii):
Let $E$ be the event of drawing a green square from the bag

We know that, Probability $P (E)$ $= \frac{(No.of favorable outcomes)}{(Total no.of possible outcomes)}$ $= \frac{9}{44}$

Therefore, the Probability that the green square is drawn $= \frac{9}{44}$

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