A piggy bank contains hundred 50 paise coins, fifty ` 1 coins, twenty ` 2 coins and ten ` 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, find the probability that the coin which fell
(i) will be a 50 paise win
(ii) will be of value more than ` 1
(iii) will be of value less than ` 5
(iv) will be a ` 1 or ` 2 coin
Answers
(i) The probability that the coin will be a 50 paise win is 5 / 9.
(ii) The probability that the coin will be of value more than 1 is 1 / 6.
(iii) The probability that the coin will be of value less than 5 is 17 / 18.
(iv) The probability that the coin will be a 1 or a 2 rupee coin is 7 / 18.
• Given,
Number of 50 paise coins = 100
Number of 1 rupee coins = 50
Number of 2 rupee coins = 20
Number of b5 rupee coins = 10
∴ Total number of coins = 100 + 50 + 20 + 10 = 180
• Probability of an event = Number of favourable outcomes / Total number of possible outcomes
(i) Probability of dropping a 50 paise coin (P₁) = Number of 50 p coins / Total number of coins
=> P₁ = 100 / 180
=> P₁ = 5 / 9
(ii) Probability of dropping a coin of value more than 1 (P₂) = (Number of 2 rs. coins + Number of 5 rs. coins) / Total number of coins
=> P₂ = (20 + 10) / 180
=> P₂ = 30 / 180
=> P₂ = 1 / 6
(iii) Probability of dropping a coin of value less than 5 (P₃) = (Number of 50 p coins + Number of 1 re. coins + Number of 2 rs. coins) / Total number of coins
=> P₃ = (100 + 50 + 20) / 180
=> P₃ = 170 / 180
=> P₃ = 17 / 18
(iv) Probability of dropping a coin of value 1 re. or 2 rs. (P₄) = (Number of 1 re. coins + Number of 2 rs. coins) / Total number of coins
=> P₄ = (50 + 20) / 180
=> P₄ = 70 / 180
=> P₄ = 7 / 18