Question

A piggy bank contains a hundred $50$ paise coins, fifty Rs. $1$ coins, twenty Rs. $2$ coins and ten Rs. $5$ 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 Rs. $1$ (iii) will be of value less than Rs. $5$ (iv) will be Rs. $1$ or Rs. $2$ coin

Answer 1

Total no. of coins in the piggy is $180$

Solution(i):

No. of $50$p coins in piggy is $100$

Let $E$ be the event of drawing a coin from piggy

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{100}{180}$$=\frac{5}{9}$

Therefore,Probability that a $50$p coin wins $=\frac{5}{9}$

Solution(ii):

No. of coins with value $>Rs.1$ in piggy is $30$

Let $E$ be the event of drawing a coin from piggy

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{30}{180}$$=\frac{1}{6}$

Therefore,Probability that a coin with value $>Rs.1$ falls $=\frac{1}{6}$

Solution(iii):

No. of coins with value $<Rs5$ in piggy is $170$

Let $E$ be the event of drawing a coin from piggy

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{170}{180}$$=\frac{17}{18}$

Therefore,Probability that a coin with value $<Rs.5$ falls$=\frac{17}{18}$

Solution(iv):

No. of Rs. $1$ and Rs. $2$ coins in piggy is $70$

Let $E$ be the event of drawing a coin from piggy

We know that, Probability $P\left(E\right)$ $=\frac{\text{(No.of favorable outcomes)}}{\text{(Total no.of possible outcomes)}}$$=\frac{70}{180}$$=\frac{7}{18}$

Therefore,Probability that a Rs. $1$ and Rs. $2$ coin falls $=\frac{7}{18}$