A) 0.55
B) 0.44
C) 0.22
D) 0.33
Correct Answer: A
Solution :
Total number of cases obtained by taking multiplication of only two numbers out of \[100={}^{100}{{C}_{2}}.\] Out of hundred \[(1,\,\,2,\,.........,\,100)\] given numbers, there are the numbers \[3,\,\,6,\,\,9,\,\,12,\,.........,\,99,\] which are 33 in number such that when any one of these is multiplied with any one of remaining 67 numbers or any two of these 33 are multiplied, then the resulting products is divisible by 3. Then the number of numbers which are the products of two of the given number are divisible by \[3={}^{33}{{C}_{1}}\times {}^{67}{{C}_{1}}+{}^{33}{{C}_{2}}.\] Hence the required probability \[=\frac{{}^{33}{{C}_{1}}\times {}^{67}{{C}_{1}}+{}^{33}{{C}_{2}}}{{}^{100}{{C}_{2}}}=\frac{2739}{4950}=0.55.\]
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