Question
Question: If \(4\) digit numbers not less than \(5000\) are randomly formed from the digits \(0,1,3,5\) and \(...
If 4 digit numbers not less than 5000 are randomly formed from the digits 0,1,3,5 and 7 , what is the probability of forming a number divisible by 5 when
the digits are repeated?
the repetition of digits is not allowed?
Solution
Here two concepts are used-
Probability of an event happening = Number of favorable cases/Number of total cases
Combination is a selection of items from a collection, such that the order of selection does not matter. Formula-
nCr=r!(n−r)!n!, where n is the number of things to choose from and we choose r from them.
In our question we will see two cases to form a number from a n number of digits to fill r number of places,
when digits are repeated=nC1×nC1×nC1×.....×rthterm=(nC1)r
when digits are not repeated=nC1×n−1C1×n−2C1×.....n−(r−1)C1
Complete answer:
(i)The number of numbers which are not less than 5000and formed by 0,1,3,5 and 7when the digits are repeated are-
We have only two digits 5 and 7 that can be at thousand’s place so that the number is not less than5000.
First we will take 5 as thousand’s place digit-
Now we have 5digits to choose from to fill 3 places so we have to choose 3 times.
Therefore the number of possible ways =53=125
Then we will take 7 as thousand’s place digit-
And again we have 5digits to choose from to fill 3 places so we have to choose 3 times.
Therefore the number of possible ways =53=125
Finally the number of numbers which are not less than 5000and formed by 0,1,3,5 and 7when the digits are repeated are =125+125=250 Statement (1)
(ii)To find the probability of the first case when digits are repeated, this would be the total number of cases.
And we have to find the number of numbers which are not less than 5000and formed by 0,1,3,5 and 7when the digits are repeated are also divisible by 5.
The divisibility rule for 5 says that the one’s place digit should be 5 or 0.
When a thousand's place digit is 5and the one’s place digit is 5.
Now we have 5digits to choose from to fill 2 places so we have to choose 2 times.
Therefore the number of possible ways =52=25
When a thousand's place digit is 5and the one’s place digit is 0.
Now we have 5digits to choose from to fill 2 places so we have to choose 2 times.
Therefore the number of possible ways =52=25
When a thousand's place digit is 7and the one’s place digit is 5.
Now we have 5digits to choose from to fill 2 places so we have to choose 2 times.
Therefore the number of possible ways =52=25
When a thousand's place digit is 7and the one’s place digit is 0.
Now we have 5digits to choose from to fill 2 places so we have to choose 2 times.
Therefore the number of possible ways =52=25
Finally the number of numbers which are not less than 5000and formed by 0,1,3,5 and 7 also divisible by 5when the digits are repeated are =25+25+25+25=100
Statement (2)
To find the probability of the first case when digits are repeated, this would be the number of favorable cases.
Probability of forming a number which are not less than 5000and formed by 0,1,3,5 and 7 also divisible by 5when the digits are repeated
= Number of favorable cases/total number of cases
=250100 [From statement (1) and statement (2)]
=52
(iii)The number of numbers which are not less than 5000and formed by 0,1,3,5 and 7when the digits are not repeated are-
We have only two digits 5 and 7 that can be at thousand’s place so that the number is not less than5000.
First we will take 5 as thousand’s place digit-
Now we have 4digits for our first choice, 3 digits for the second and 2 for the third.
Therefore the number of possible ways here=4×3×2=24
Then we will take 7 as thousand’s place digit-
Now we have 4digits for our first choice, 3 digits for the second and 2 for the third.
Therefore the number of possible ways here=4×3×2=24
Finally the number of numbers which are not less than 5000and formed by 0,1,3,5 and 7when the digits are not repeated are =24+24=48 Statement (3)
To find the probability of a second case when digits are not repeated, this would be the total number of cases.
And we have to find the number of numbers which are not less than 5000and formed by 0,1,3,5 and 7when the digits are not repeated are also divisible by 5.
The divisibility rule for 5 says that the one’s place digit should be 5 or0.
When a thousand's place digit is 5and the one’s place digit is5.
This case cannot be possible since 5is repeated.
When a thousand's place digit is 5and the one’s place digit is0.
Now we have 3digits for our first choice and 2 digits for the second.
Therefore the number of possible ways here=3×2=6
When a thousand's place digit is 7and the one’s place digit is 5.
Now we have 3digits for our first choice and 2 digits for the second.
Therefore the number of possible ways here=3×2=6
When a thousand's place digit is 7and the one’s place digit is 0.
Now we have 3digits for our first choice and 2 digits for the second.
Therefore the number of possible ways here=3×2=6
Finally the number of numbers which are not less than 5000and formed by 0,1,3,5 and 7 also divisible by 5when the digits are not repeated are =6+6+6=18 Statement (4)
To find the probability of a second case when digits are repeated, this would be the number of favorable cases.
Probability of forming a number which are not less than 5000and formed by 0,1,3,5 and 7 also divisible by 5when the digits are not repeated
= Number of favorable cases/total number of cases
=4818 [from statement (3) and statement (4)]
=83
Hence,
Probability of forming a number which are not less than 5000and formed by 0,1,3,5 and 7 also divisible by 5when the digits are repeated =52
Probability of forming a number which are not less than 5000and formed by 0,1,3,5 and 7 also divisible by 5when the digits are not repeated =83
Note: The number of ways to arrange n objects in a row, of which exactly r objects are identical, isr!n!.
In generalize form, if we wish to arrange a total no. of nobjects, out of which p are of one type, q of second type are alike, and r of a third kind are same, then such a computation is done by p!q!r!n!