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JEE Advanced · Mathematics · 4. P&C

Let X be the set of all five digit numbers formed using 1,2,2,2,4,4,0. For example, 22240 is in X while 02244 and 44422 are not in X. Suppose that each element of X has an equal chance of being chosen. Let p be the conditional probability that an element chosen at random is a multiple of 20 given that it is a multiple of 5. Then the value of 38 p is equal to

  1. A 31
  2. B 32
  3. C 33
  4. D 34
Verified Solution

Answer & Solution

Correct Answer

(A) 31

Step-by-step Solution

Detailed explanation

First we will find the sample space in which the number of five-digit numbers are divisible by 5,
So, fixing zero at the last place we get,
----0
Now in first four place following number can take place,
2224 4!3!=4 ways
22444!2!2!=6 ways
22214!3!=4 ways
22414!2!=12 ways
24414!2!=12 ways
So, total sample space will be 4+6+4+12+12=38 
Now finding the number of favourable outcomes,
So, Number of five-digit numbers divisible by 5 but 'not' by 20
Now fixing 10 in last two places, we get
---1 0
So, the first three places can be occupied by,
222  1 ways
224 3 ways
244 3 ways
So, total number of numbers which are divisible by 5 but not 20 will be, 1+3+3=7
So, favourable number of five-digit numbers that are divisible by 5 and 20=38-7=31
Hence, probability is given by, p=3138
38 p=31
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