What is probability (approximate) of a 4 digit cube root random number .
a .0004
b .0005
c .0011
d . 00331

• 1000,1331,1728,2197 .......4000,9261,
  (10^3,11^3,12^3............20^3,21^3)
  total possibilities are 12
  so probability is 12/9000(total number 4 digit nubers are 9000(9999-999)
  answer is .00133
  
• 10 years agoHelpfull: Yes(44) No(0)
• Please bro first write all option correctly. The right answer is- 12/9000= .00133..
• 10 years agoHelpfull: Yes(16) No(0)
• 1. first count how many 4digit numbers are threre?? (1000 to 9999)=9000(including 1000 and 9999)
  2. now take cube root of 1000=10^3, likely as 11^3, 12^3, 13^3, 14^3 up to 9261=21^3(extreme 4 digit cube root)
  3. so 10,11,12,13 up to 21 total numerics=12 (including 10 and 21).
  4. now probability
  
  P(N)=n(E) / n(S)
  =12 /9000
  = 0.0013333 (given Options are all wrong) -Thank you
• 9 years agoHelpfull: Yes(5) No(0)
• I am not getting the question plz explain once
• 10 years agoHelpfull: Yes(2) No(4)
• The answer might be 12/32...
• 9 years agoHelpfull: Yes(0) No(1)