Question

Question 9 If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when,

(i) the digits are repeated?

(ii) the repetition of digits is not allowed?

Class X1 - Maths -Probability Page 409

Answer 1

case 1 :- when digits are repeated
Total four digits number which are greater than 5000 . from the digits 0, 1, 3 , 5 and 7 :

In thousand place , the number will be 5 or 7 , and rest of three places any five number will come.
hence, thousand place can be filled in 2 ways
hundred place can be filled in 5 ways { because repeatation is allowed }
tenth place can be filled in 5 ways
and unit place can be filled in 5 ways .

so, total four digit numbers n ( S ) = 2 × 5 × 5 × 5 = 250

now, Total four digits numbers which are divisible by 5 .
e.g in unit place will be 0 or 5 and thousand place will be 5 or 7 and rest two places any five numbers will be come .
so,
thousand place can be filled = 2 ways
hundred place can be filled = 5 ways
tenth place can be filled = 5 ways
unit place can be filled = 2 ways
so, total four digits numbers which are divisible by 5 n( E ) = 2 × 5 × 5 × 2 = 100

so, P( E ) = n ( E )/n ( S ) = 100/250 = 2/5


Case 2:- when repetition are not allowed.
Total four digits numbers : for greater than 5000 , thousand place can be filled with 5 or 7 . since , the repetition are not allowed , then rest of the place can be filled by 4 , 3 , 2

hence, total four digits numbers n( S ) = 2 × 4 × 3 × 2 = 48 ways

total four digits numbers which are divisible by 5 :
when thousand place filled with 5 and unit place filled with 0 then , rest two palces may be filled by 3, 2 ways .so, no of ways = 6

when thousand place filled by 7 and unit palace filled by 0 then, rest of two places may be filled by 3 , 2 ways . so, no of ways = 6

when thousand place filled by 7 and unit place filled by 5 then rest of two places may be filled by 3, 2 ways . so, no of ways = 6

so, total four digits numbers which are divisible by 5 n ( E ) = 6 + 6 + 6 = 18

so, P( E ) = n( E )/n( S ) = 18/48 = 3/8

Answer 2

Answer:

Step-by-step explanation:

A 4 digit number greater than 5000 is randomly formed from digits 0,1,3,5,7.

(1) Repetition is allowed:

We need to form a number greater than 5000, hence, the leftmost digit can be either 5 or 7.

Since, repetition of digits is allowed, so the remaining three places can be filled by 0, 1, 3, 5, or 7.

Hence, the total number of 4 digit numbers that can be formed greater than 5000 are = 2.5.5.5 = 250

But, we can’t count 5000 so the total number becomes 250 – 1 = 249.

The number is divisible by 5 only if the numnber at unit’s place is either 0 or 5.

Hence, the total number of numbers greater than 5000 and divisible by 5 are = 2.5.5.2 – 1 = 99

Hence, the required probability is given by = 99/249

= 33/83.

(2) If repetition of digits is not allowed:

For a number to be greter than 5000, the digit at thousand’s place can be either 5 or 7.

The remaining three places can be filled by any of the four digits.

hence, total number of numbers greater than 5000 = 2.4.3.2 = 48.

When the digit at thousand’s place is 5, units digit can be 0 and the tens and hundreds digit can be any two of the remaining three digits.

Hence, the number of 4 digit numbers starting with 5 and divisible by 5 = 3.2 = 6

When the digit at thousand’s place is 7, units digit can be filled in two ways (0 or 5) and the tens and hundreds digit can be any two of the remaining three digits.

Hence, the number of 4 digit numbers starting with 7 and divisible by 5 = 1.2.3.2 = 12.

therefore, the number of 4 digit numbers greater than 5000 and divisible by 5 = 12 + 6 = 18

hence, the required probability = 18/48 = 3/8.