
A piggy bank contains a hundred 50 paise coins, fifty Rs. 1 coin, 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
(a). will be a 50 paise win
(b). will be of value more than Rs. 1
(c). will be of value less than Rs. 5
(d). will be Rs. 1 or Rs. 2 coin
Answer
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- Hint: In this question, the total number of coins in the piggy bank are 180. The total number of coins of each denomination is given in the problem statement. By using the data, we calculate the favourable outcomes of each event. After that by using the formula of probability of an event we can easily calculate all the parts of our question.
Complete step-by-step solution -
Formula of probability used in our question:
$\text{P}\left( \text{Event} \right)\text{=}\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
The total number of coins in the piggy bank = number of coins of 50 paise + number of coins of 1 rupee + number of coins of 2 rupees + number of coins are 5 rupees = 100 + 50 +20 +10 = 180 coins.
(a). Number of 50 paise coins = 100
Total number of coins = 180.
Now, by using the above stated formula, we get
$\begin{align}
& P(Event)=\dfrac{100}{180} \\
& P(Event)=\dfrac{5}{9} \\
\end{align}$
Hence, the probability that a 50 paise win is $\dfrac{5}{9}$.
(b). Number of coins with value greater than Rs. 1 = 30.
Now, by using the above stated formula, we get
$\begin{align}
& P(Event)=\dfrac{30}{180} \\
& P(Event)=\dfrac{1}{6} \\
\end{align}$
Hence, the probability of value more than Rs. 1 are $\dfrac{1}{6}$.
(c). Number of coins with value less than Rs. 5 in piggy = 170.
Now, by using the above stated formula, we get
$\begin{align}
& P(Event)=\dfrac{170}{180} \\
& P(Event)=\dfrac{17}{18} \\
\end{align}$
Hence, the probability of value less than Rs. 5 is $\dfrac{17}{18}$.
(d).Number of coins of 1 rupee and 2 rupees = 50 + 20 =70.
Now, by using the above stated formula, we get
$\begin{align}
& P(Event)=\dfrac{70}{180} \\
& P(Event)=\dfrac{7}{18} \\
\end{align}$
Hence, the probability of a Rs. 1 or Rs. 2 coin is $\dfrac{7}{18}$.
Note: The key concept involved in solving this problem is the knowledge of probability of an event. Students must be careful while calculating the favourable outcomes according to the mentioned case. Redundancy of events should be avoided while calculating total outcomes.
Complete step-by-step solution -
Formula of probability used in our question:
$\text{P}\left( \text{Event} \right)\text{=}\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$
The total number of coins in the piggy bank = number of coins of 50 paise + number of coins of 1 rupee + number of coins of 2 rupees + number of coins are 5 rupees = 100 + 50 +20 +10 = 180 coins.
(a). Number of 50 paise coins = 100
Total number of coins = 180.
Now, by using the above stated formula, we get
$\begin{align}
& P(Event)=\dfrac{100}{180} \\
& P(Event)=\dfrac{5}{9} \\
\end{align}$
Hence, the probability that a 50 paise win is $\dfrac{5}{9}$.
(b). Number of coins with value greater than Rs. 1 = 30.
Now, by using the above stated formula, we get
$\begin{align}
& P(Event)=\dfrac{30}{180} \\
& P(Event)=\dfrac{1}{6} \\
\end{align}$
Hence, the probability of value more than Rs. 1 are $\dfrac{1}{6}$.
(c). Number of coins with value less than Rs. 5 in piggy = 170.
Now, by using the above stated formula, we get
$\begin{align}
& P(Event)=\dfrac{170}{180} \\
& P(Event)=\dfrac{17}{18} \\
\end{align}$
Hence, the probability of value less than Rs. 5 is $\dfrac{17}{18}$.
(d).Number of coins of 1 rupee and 2 rupees = 50 + 20 =70.
Now, by using the above stated formula, we get
$\begin{align}
& P(Event)=\dfrac{70}{180} \\
& P(Event)=\dfrac{7}{18} \\
\end{align}$
Hence, the probability of a Rs. 1 or Rs. 2 coin is $\dfrac{7}{18}$.
Note: The key concept involved in solving this problem is the knowledge of probability of an event. Students must be careful while calculating the favourable outcomes according to the mentioned case. Redundancy of events should be avoided while calculating total outcomes.
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