Question

Using the digits \[2,4\] and \[6,\] all possible \[2 - \]digit numbers are formed, each digit used only once. Find the probability that the number so formed is greater than \[46.\]

Answer 1

Hint: To solve this question, we will start with finding total outcomes of \[2 - \] digit number using \[2,4\] and \[6,\] and the number of favourable outcomes i.e., number which are greater than \[46.\] Then we will use the probability formula and get the required answer.

Complete step by step answer:
We have been given that by using only the digits \[2,4\] and \[6,\] all possible \[2 - \]digit numbers are formed, where each digit is used only once. And we need to find the probability that the number so formed is greater than \[46.\]

So, total outcomes of all possible \[2 - \]digit numbers formed using the digits \[2,4\] and \[6\] \[ = {\text{ }}\left\{ {24,{\text{ }}26,{\text{ }}42,{\text{ }}46,{\text{ }}62,{\text{ }}64} \right\}{\text{ }} = {\text{ }}6\]

And, number of favourable outcomes of \[2 - \]digit numbers which are greater than \[46{\text{ }} = {\text{ }}\left\{ {62,{\text{ }}64} \right\}{\text{ }} = {\text{ }}2\]

We, know that, Probability $ = \dfrac{{Favourable{\text{ }}outcomes}}{{Total{\text{ }}outcomes}}$

Now, putting the values in the formula mentioned above, we get
Probability of getting number greater than \[46\]$ = \dfrac{2}{6} = \dfrac{1}{3}$
Thus, the probability of getting \[2 - \]digit number greater than \[46,\] using the digits \[2,4\] and \[6\] is $\dfrac{1}{3}$.

Note: To solve this question, students should carefully write down the \[2 - \] digit numbers that can be formed using the digits given. As this should be the first step taken by you and any mistake in it can make your answer wrong.