
In a lottery of 50 tickets numbered from 1 to 50, one ticket is drawn. Find the probability that the drawn ticket bears a prime number.
Answer
577.2k+ views
Hint: To solve this question first we need to list all the favorable outcomes by calculating the prime numbers from $1$ to $50$. Then we count the total number of outcomes as lottery tickets are numbered from 1 to 50 and use the formula to calculate the probability. Following formula is used-
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Where, A is an event,
\[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes
Complete step-by-step solution:
We have been given that in a lottery of 50 tickets numbered from 1 to 50, one ticket is drawn.
We have to find the probability that the drawn ticket bears a prime number.
Now, we know that the probability of an event is given by the formula
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Where, A is an event,
\[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes
Now, we have 50 tickets numbered from 1 to 50, so total number of possible outcomes will be $n\left( S \right)=50$
Now, we have $15$ prime numbers from 1 to 50.
So, the number of favorable outcomes will be \[n\left( E \right)=15\]
Now, the probability that the one drawn ticket bears a prime number will be
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
$\begin{align}
& P\left( A \right)=\dfrac{15}{50} \\
& P\left( A \right)=\dfrac{3}{10} \\
\end{align}$
So, the probability that the one drawn ticket bears a prime number is $\dfrac{3}{10}$.
Note: A prime number has only two distinct factors, one and the number itself. $2$ is the only prime number which is even, all other prime numbers are odd numbers. The number of $1$ is not a prime number. Some students may count $1$ also in the list of prime numbers and get an incorrect answer. The list of prime numbers from 1 to 50 is $2,3,5,7,11,13,17,19,23,29,31,37,41,43,47$.
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Where, A is an event,
\[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes
Complete step-by-step solution:
We have been given that in a lottery of 50 tickets numbered from 1 to 50, one ticket is drawn.
We have to find the probability that the drawn ticket bears a prime number.
Now, we know that the probability of an event is given by the formula
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Where, A is an event,
\[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes
Now, we have 50 tickets numbered from 1 to 50, so total number of possible outcomes will be $n\left( S \right)=50$
Now, we have $15$ prime numbers from 1 to 50.
So, the number of favorable outcomes will be \[n\left( E \right)=15\]
Now, the probability that the one drawn ticket bears a prime number will be
$P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
$\begin{align}
& P\left( A \right)=\dfrac{15}{50} \\
& P\left( A \right)=\dfrac{3}{10} \\
\end{align}$
So, the probability that the one drawn ticket bears a prime number is $\dfrac{3}{10}$.
Note: A prime number has only two distinct factors, one and the number itself. $2$ is the only prime number which is even, all other prime numbers are odd numbers. The number of $1$ is not a prime number. Some students may count $1$ also in the list of prime numbers and get an incorrect answer. The list of prime numbers from 1 to 50 is $2,3,5,7,11,13,17,19,23,29,31,37,41,43,47$.
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