Question

The two numbers 'a' and 'b' are selected successively such that b-a = 1 in that order from the integers 1 to 10. The probability that b/a
is an integer is ​

Answer 1

$Given,two numbers are selected without replacement from integers 1 to 10.</p><p></p><p></p><p>Now, favourable cases for which ba will be an integer are:</p><p></p><p></p><p>12=2;13=3;14=4;24=2;15=5;</p><p></p><p></p><p>16=6;26=3;36=2;17=7;18=8;</p><p></p><p></p><p>28=4;48=2;19=9;39=3;110=10;</p><p></p><p></p><p>210=5;510=2</p><p></p><p></p><p>∴, favourable number of cases=17</p><p></p><p></p><p>Total cases are:</p><p></p><p></p><p>(21,31,...101),(12,32,...,102),...,(110,210,310,...,910)</p><p></p><p></p><p>So, total number of cases=9×10=90</p><p></p><p></p><p>∴ required probability=9017</p><p></p><p></p><p>$

Answer 2

Given,two numbers are selected without replacement from integers 1 to . Now, favourable cases for which will be an integer are: ;;;;; ;;;;; ;;;;; ; , favourable number of cases Total cases are: So, total number of cases required probability