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Home  >>  CBSE XI  >>  Math  >>  Probability
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If four digit numbers greater than 5000 are randomly formed from the digits 0,1,3,5 and 7,what is the probability of forming number divisible by 5 when the digits are repeated ?

$\begin{array}{1 1}(A)\;\large\frac{2}{5}\\(B)\;\large\frac{3}{5}\\(C)\;\large\frac{4}{5}\\(D)\;\large\frac{7}{5}\end{array} $

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1 Answer

–1 vote
  • Required probability =$\large\frac{n(E)}{n(S)}$
Step 1:
4 digit number greater than 5000 are to be formed
Digits -0,1,3,5,7
Number to be divisible by 5
$\therefore$ A number can be divisible by 5 only when its units digit is 0,5
Step 2:
The digits are repeated
Total number of outcomes will be :
Since the number has to be greater than 5000,the thousand digit should be 5 or 7 and rest three digit can be any of the number.Total number of digits =5
$\therefore 5\Rightarrow 5\times 5\times 5=125$
$7\Rightarrow 5\times 5\times \times 5=125$
$\therefore$ Total 4 digit numbers n(S)=250
Step 3:
Number of favorable outcomes :
For a number to be divisible by 5.The unit digit should be 0 or 5 and the thousand digit is fixed as 5 or 7.
$\therefore$ Total number of favorable outcomes n(E)=25+25+25+25
$\Rightarrow 100$
Step 4 :
$\therefore$ Required probability =$\large\frac{n(E)}{n(S)}$
$\Rightarrow \large\frac{100}{250}$
$\Rightarrow \large\frac{2}{5}$
Hence (A) is the correct answer.
answered Jul 3, 2014 by sreemathi.v
Question is for number greater than 5000
So, Total No. of Outcomes will be 250-1=249
and No. of Favorable Events will be 100-1=99
Therefore, Probability is 99/249 = 33/83

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