A piggy bank contains hundred 50-p coins, seventy R s. 1 coin and thir

1.	A piggy bank contains hundred 50-p coins, seventy R s. 1 coin and thirty R s. 5 coins. If it is equally likely one of the coins will fall out when the blank is turned upside down, what is the probability that the coin(i) Will be a Rs. 1 coins ?(ii) Will not be a Rs. 5 coins(iii) Will not 50-p or a Rs. 2 coin ?
Answer» Number of 50-p coins = 100. Number of 1 coins = 70. Number of 2 coins = 50. Number of 5 coins = 30. Thus, total number of outcomes = 250. (i) Let E₁ be the event of getting a Rs. 1 coin. The number of favorable outcomes = 70. Therefore, P(getting a Rs. 1 coin) = P(E₁) = \(\frac{number\, of \,outcomes\,favorable\,to\,E_1}{number\,of \,all\,possible\,outcomes} \) = \(\frac{70}{250}\) = \(\frac{7}{25}\) Thus, the probability that the coin will be a Rs. 1 coin is \(\frac{7}{25}\) (ii) Let E₂be the event of not getting a Rs. 5 Number of favorable outcomes = 250 - 30 = 220 Therefore, P( not getting a Rs. 5 coin) = P(E₂) = \(\frac{number\, of \,outcomes\,favorable\,to\,E_2}{number\,of \,all\,possible\,outcomes} \) = \(\frac{220}{250}\) = \(\frac{22}{25}\) Thus, probability that the coin will not be a Rs. 5 coin is \(\frac{22}{25}\). (iii) Let E₃be the event of getting a 50-p or a Rs 2 coins. Number of favorable outcomes = 100 + 50 = 150 Therefore, P(getting a a 50-p or a Rs 2 coin) P(E₃) = \(\frac{number\, of \,outcomes\,favorable\,to\,E_3}{number\,of \,all\,possible\,outcomes} \) = \(\frac{150}{250}\) = \(\frac{3}{5}\) Thus, probability that the coin will be a 50-p or a Rs 2 coin is \(\frac{3}{5}\).

A piggy bank contains hundred 50-p coins, seventy R s. 1 coin and thirty R s. 5 coins. If it is equally likely one of the coins will fall out when the blank is turned upside down, what is the probability that the coin(i) Will be a Rs. 1 coins ?(ii) Will not be a Rs. 5 coins(iii) Will not 50-p or a Rs. 2 coin ?

Answer»

Number of 50-p coins = 100.

Number of 1 coins = 70.

Number of 2 coins = 50.

Number of 5 coins = 30.

Thus, total number of outcomes = 250.

(i) Let E₁ be the event of getting a Rs. 1 coin.

The number of favorable outcomes = 70.

Therefore, P(getting a Rs. 1 coin) = P(E₁) = \(\frac{number\, of \,outcomes\,favorable\,to\,E_1}{number\,of \,all\,possible\,outcomes} \) = \(\frac{70}{250}\) = \(\frac{7}{25}\)

Thus, the probability that the coin will be a Rs. 1 coin is \(\frac{7}{25}\)

(ii) Let E₂be the event of not getting a Rs. 5

Number of favorable outcomes = 250 - 30 = 220

Therefore, P( not getting a Rs. 5 coin) = P(E₂) = \(\frac{number\, of \,outcomes\,favorable\,to\,E_2}{number\,of \,all\,possible\,outcomes} \) = \(\frac{220}{250}\) = \(\frac{22}{25}\)

Thus, probability that the coin will not be a Rs. 5 coin is \(\frac{22}{25}\).

(iii) Let E₃be the event of getting a 50-p or a Rs 2 coins.

Number of favorable outcomes = 100 + 50 = 150

Therefore, P(getting a a 50-p or a Rs 2 coin) P(E₃) = \(\frac{number\, of \,outcomes\,favorable\,to\,E_3}{number\,of \,all\,possible\,outcomes} \) = \(\frac{150}{250}\) = \(\frac{3}{5}\)

Thus, probability that the coin will be a 50-p or a Rs 2 coin is \(\frac{3}{5}\).

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