A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 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 Re 1 coin? (ii) will not be a Rs 2 coin?
A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 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 Re 1 coin? (ii) will not be a Rs 2 coin?

Solution:

We have,

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

So, the total possible outcomes = 100 = n(S)

(i) Number of favourable outcomes for Re 1 coins = 30 = n(E)

Probability (Re 1 coin) = n(E)/ n(S) = 30/100 = 3/10

(ii) Number of favourable outcomes for not a Rs 2 coins = number of favourable outcomes for Re 1 or Rs 5 coins = 30 + 20 = 50 = n(E)

Hence, probability (not Rs 2 coin) = n(E)/ n(S) = ½