A bag contains twenty ₹ 5 coins, fifty ₹ 2 coins and thirty ₹ 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin :

(i) will be a ₹ 1 coin ?

(ii) will not be a ₹ 2 coin ?

(iii) will neither be a ₹ 5 coin nor be a ₹ 1 coin ?

We have,

Total number of coins = 20 + 50 + 30 = 100

So, the total possible outcomes = 100.

(i) Number of ₹ 1 coin = 30

∴ Number of favourable outcomes = 30

P(drawing a ₹ 1 coin) = $\dfrac{\text{No. of favourable outcomes}}{\text{\text{No. of possible outcomes}}} = \dfrac{30}{100} = \dfrac{3}{10}$.

Hence, the probability of drawing a ₹ 1 coin = $\dfrac{3}{10}$.

(ii) Number of ₹ 5 and ₹ 1 coin = 50 (20 + 30)

Hence, no. of coins apart from ₹ 2 coins = 50.

∴ Number of favourable outcomes = 50

P(coin drawn will not be a ₹ 2 coin) = $\dfrac{\text{No. of favourable outcomes}}{\text{\text{No. of possible outcomes}}} = \dfrac{50}{100} = \dfrac{1}{2}$.

Hence, the probability of not drawing a ₹ 2 coin = $\dfrac{1}{2}$.

(iii) No. of ₹ 2 coins = 50

No. of favourable outcomes (for drawing a ₹ 2 coin) = 50

P(drawing a ₹ 2 coin) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of total possible outcomes}} = \dfrac{50}{100} = \dfrac{1}{2}$.

Since, there are only 3 types of coins in the bag.

∴ P(drawing neither ₹ 5 nor ₹ 1 coin) = P(drawing ₹ 2 coin) = $\dfrac{1}{2}$.

Hence, the probability of drawing neither ₹ 5 nor ₹ 1 coin = $\dfrac{1}{2}$.