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
A. A
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 2244 → 4 ! 2 ! 2 ! = 6   ways 2221 → 4 ! 3 ! = 4   ways 2241 → 4 ! 2 ! = 12   ways 2441 → 4 ! 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 = 31 38 ⇒ 38   p = 31
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