Answer
$Each\:digit\:\left(0-9\right)\:has\:an\:equal\:probability\:of\:occurring\:on\:average,$
$\:though\:not\:necessarily\:in\:every\:sequence.\:Over\:many\:sequences,\:each\:$
$digit\:should\:appear\:roughly\:once\:in\:every\:10\:digits.$
Work Step by Step
$The\:theory\:behind\:random\:numbers\:is\:that\:each\:digit,\:0\:through\:9,\:$
$has\:an\:equal\:probability\:of\:occurring.\:This\:does\:not\:mean\:that\:in\:$
$every\:sequence\:of\:10\:digits\:has\:a\:\frac{1}{10}\:probability.\:This\:does\:not\:mean\:$
$that\:in\:every\:sequence\:of\:10\:digits,\:you\:will\:find\:each\:10\:digit.\:Rather,\:it\:$
$means\:that\:on\:the\:average,\:each\:digit\:will\:occur\:once.\:For\:example,\:$
$the\:digit\:2\:may\:occur\:3\:times\:in\:a\:sequence\:of\:10\:digits.\:But\:in\:later\:$
$sequences,\:it\:may\:not\:occur\:at\:all\:and\:thus\:averaging\:to\:be\:a\:probability\:$
$of\:\frac{1}{10}.$
