Question

If $4$ digit number not less than $5000$ are randomly formed from the digits $0,1,3,5{\text{ and }}7$ , what is the probability of forming a number divisible by $5$ when:
(i) the digits are repeated?
(ii) the repetition of a digit is not allowed?

Answer 1

Since the number should be greater than $5000$ , the only possible digits for thousands place is $5{\text{ and 7}}$. For the case when the digits are not repeated, you figure out the number of choices on each position and then multiply them to find the total number of arrangements. For the second case, make cases with numbers of form $5{\text{XXX}}0,7{\text{XXX}}0{\text{ and 7XXX5}}$ and calculate the possible choices for the middle three positions. Find favourable outcomes and total outcomes and take their ratio for finding probability.

Complete step-by-step answer:
Let’s start with analysing the problem here. We need to find the probability of selecting a number divisible by five from a group of four-digit numbers formed using digits $0,1,3,5{\text{ and 7}}$ and is not less than $5000$, i.e. greater than or equal to $5000$ .
We need to find this probability as per the two cases (i) and (ii).
Before moving forward with the solution we should understand the concept of probability. Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. how likely they are to happen, using it. Probability can range from $0$ to $1$ , where $0$ means the event to be an impossible one and $1$ indicates a certain event.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
$\Rightarrow {\text{Probability of event to happen P(E) = }}\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}$
Considering Case (i)
Here, we need to figure out the probability of forming a number divisible by five when the numbers are formed with repetition being allowed in the digits.
As per the formula of probability, let’s find the number of favourable outcomes and the total number of outcomes separately.
So, the total number of four-digit numbers greater than $5000$ that can be formed using $0,1,3,5{\text{ and 7}}$ with repetition can be found by fixing $5{\text{ and 7}}$ at the thousands place and placing any of the five digits in the remaining three places.
Therefore, we have $2$ digits for thousands place, $5$ for hundreds place, $5$ for tens place and also $5$ for one's place $\left( {{\text{Any one from 0,1,3,5 or 7}}} \right)$ .
As we know that the number of ways for the arrangement of $m{\text{ and }}n$ things simultaneously can be given by the product $\left( {m \times n} \right)$
$\Rightarrow {\text{Total number of outcomes }} = 2 \times 5 \times 5 \times 5 = 250$
For the number of favourable outcomes, we need to fix $5{\text{ and 7}}$ at thousands place and also $0{\text{ or 5}}$ at the ones place since the multiples of five always have either five or zero at units place.
So, here we have $2$ choices for thousands place, $5$ choices for hundreds place, $5$ choices for tens place and $2$ choices for units place.
$\Rightarrow$ Number of favourable outcomes $= 2 \times 5 \times 5 \times 2 = 100$
Now for finding the required probability we can use the formula for probability and substitute these values in it.
Therefore, the required probability$= \dfrac{{100}}{{250}} = \dfrac{2}{5} = 0.4$
Considering case (ii)
Here, we need to figure out the probability of a number being divisible by five, selected from a group of a four-digit number greater than $5000$ formed using the digits $0,1,3,5{\text{ and 7}}$ with no repetition allowed in the digits.
So, the total number of possible outcomes can be calculated by fixing $5{\text{ or 7}}$ at the thousands place and then we have four remaining choices for the hundreds place, then three choices for tens place and two for units place.
$\Rightarrow {\text{Total number of outcomes = 2}} \times 4 \times 3 \times 2 = 48$
Now for the favourable outcome, we need to find the number of four-digit numbers greater than $5000$ that can be formed using the digits $0,1,3,5{\text{ and 7}}$with no repetition and are exactly divisible by five.
As we know that for a number to be divisible by five, it should have either $0{\text{ or 5}}$ as its units digit. Therefore when we fix ${\text{7}}$ at the thousands place then we have $0{\text{ or 5}}$ as the choices for ones place. But when we fix $5$ at the thousands place then there is only one choice for ones place, i.e. the digit $0$ .
$\Rightarrow$ Number of ways of the arrangement when $5$ is fixed at thousands place and $0$ at ones place (remaining three choices for hundreds place and two for tens place)$= 3 \times 2 = 6$
$\Rightarrow$ Number of ways of the arrangement when $7$ is fixed at thousands place and $0$ at ones place (remaining three choices for hundreds place and two for tens place)$= 3 \times 2 = 6$
$\Rightarrow$Number of ways of the arrangement when $7$ is fixed at thousands place and $5$ at ones place (remaining three choices for hundreds place and two for tens place)$= 3 \times 2 = 6$
Since these arrangements are carried out separately and do not depend on others, we will add these three to get the total number of favourable outcomes.
Therefore, total favourable outcomes $= 6 + 6 + 6 = 18$
Now for finding the required probability we can use the formula for probability and substitute these values in it.
Therefore, the required probability$= \dfrac{{18}}{{48}} = \dfrac{3}{8} = 0.375$

Note: In a question like this, always try to go step by step because it helps in avoiding skipping arrangements. We used the techniques of fixing a few digits at specific places in a four-digit number. When a digit is fixed at a position and repetition is not allowed, then our number of choices gets reduced by one. This helps in satisfying the condition given in the question and makes arrangement easier.