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If 4-digit numbers greater than $\mathbf{5 , 0 0 0}$ are randomly formed from the digits $\mathbf{0}, \mathbf{1}, \mathbf{3}, 5$ 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?
(i) the digits are repeated ?
(ii) the repetition of digits is not allowed?
Solution:
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Verified Answer
(i) When digits are repeated. Let 4 places in a 4-digit number be named as
If place I is filled up with 5 , the places II, III, IV may be filled up in $5 \times 5 \times 5=125$ ways
Similarly, when place I is filled up with 7 , again place II, III, IV may be filed up in $5 \times 5 \times 5=125$ ways. When 4-digit number is greater than 5000 , the number of exhaustive cases $=125+125=250$. A number is divisible by 5 in four cases
In each case, II and III place may be filled up in $5 \times 5=25$ cases.
$\therefore$ Number of numbers which are divisible by $5=$ $4 \times 25=100$
$\therefore$ When repetition of digits is allowed. Probability that number is greater than 5000 and divisible by $5=\frac{100}{250}=\frac{2}{5}$
(ii) When digits are not repeated.
Let place I be filled up with 5. Place II, III, IV may be filled up in $4 \times 3 \times 2=24$ ways. Similarly, when 7 is placed at I, again II, III, IV places may be filled up in $4 \times 3 \times 2=24$
Total number of cases when 4-digit number is greater than $5000=24+24=48$
Now, a number is divisible by 5 is obtained when 0 or 5 is placed at IV as
In each case II and III places can be filled up in $3 \times 2=6$ ways.
When digits are not repeated the number of numbers divisible by 5 are $6+6+6=18$
$\therefore$ Probability that 4-digits number greater than 5000 and divisible by $5=\frac{18}{48}=\frac{3}{8}$
If place I is filled up with 5 , the places II, III, IV may be filled up in $5 \times 5 \times 5=125$ ways
Similarly, when place I is filled up with 7 , again place II, III, IV may be filed up in $5 \times 5 \times 5=125$ ways. When 4-digit number is greater than 5000 , the number of exhaustive cases $=125+125=250$. A number is divisible by 5 in four cases
In each case, II and III place may be filled up in $5 \times 5=25$ cases.
$\therefore$ Number of numbers which are divisible by $5=$ $4 \times 25=100$
$\therefore$ When repetition of digits is allowed. Probability that number is greater than 5000 and divisible by $5=\frac{100}{250}=\frac{2}{5}$
(ii) When digits are not repeated.
Let place I be filled up with 5. Place II, III, IV may be filled up in $4 \times 3 \times 2=24$ ways. Similarly, when 7 is placed at I, again II, III, IV places may be filled up in $4 \times 3 \times 2=24$
Total number of cases when 4-digit number is greater than $5000=24+24=48$
Now, a number is divisible by 5 is obtained when 0 or 5 is placed at IV as
In each case II and III places can be filled up in $3 \times 2=6$ ways.
When digits are not repeated the number of numbers divisible by 5 are $6+6+6=18$
$\therefore$ Probability that 4-digits number greater than 5000 and divisible by $5=\frac{18}{48}=\frac{3}{8}$
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