A box contains cards bearing numbers 6 to 70. If one card is drawn at random from the box, find the probability that it bears a composite number between 50 and 70.

Hint: First find the sample space of the set of the cards by calculating the numbers from 6 to 70. Then find the number of favourable outcomes by calculating the composite numbers from 50 to 70. Determine the required probability by the formula, $P=\dfrac{Favourable\text{ outcomes}}{number \, of\, outcomes\, in \,Sample\text{ space}}$

Complete step by step solution:
Sample space, S =total number of cards from 6 to 70
Applying general term formula of an arithmetic progression;
First term, a=6
Last term, l=70
Common difference, d=1 (as consecutive numbers from 6 to 70 are taken)
Let, total number of cards is ‘n’
By the formula of last term of AP;
\[\begin{align}
  & l=a+(n-1)d \\
 & \Rightarrow 70=6+(n-1)(1) \\
 & \Rightarrow n=65 \\
\end{align}\]
No of cards from 6 to 70 is 65.
So, sample space, \[\left| S \right|=65\]
Favourable outcome, A = {a composite number between 50 and 70}
So, A = {51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69}
The number of favourable outcomes, $\left| A \right|=15$
Probability of getting a composite number between 50 and 70,
$P\left( A \right)=\dfrac{\left| A \right|}{\left| S \right|}=\dfrac{15}{65}=\dfrac{3}{13}$
This is the required solution of the given question.

Note:
In mathematics, composite numbers are the numbers which have more than two factors. So, the composite numbers between 50 and 70 are 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68 and 69. Hence, we are getting these as our favourable outcome for ‘A’.