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Real roots from three dice (Posted on 2016-09-20) Difficulty: 3 of 5
Die 1 is an unbiased and regular six-sided die numbered 1 to 6.
Die 2 is an unbiased and regular seven-sided die numbered 2 to 8.
Die 3 is an unbiased and regular eight sided die numbered 3 to 10

Consider the quadratic equation Ax2+Bx+C= 0.

We assign values to the coefficient A by throwing Die 1, the coefficient B by throwing Die 2 and the coefficient C by throwing Die 3.

Determine the probability that the equation will have real roots.

No Solution Yet Submitted by K Sengupta    
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Solution computer-aided solution Comment 1 of 1

There are 52 cases of real roots out of the 336 possible combinations that are possible.  (5 of the 52 cases are examples of double real roots)

The probability is therefore 52/336 = 13/84 ~= 0.154761904761905.

Private Sub Form_Load()
 Form1.Visible = True
 
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 For a = 1 To 6
 For b = 2 To 8
 For c = 3 To 10
   disc = b * b - 4 * a * c
   If disc >= 0 Then
     If disc = 0 Then dblCt = dblCt + 1
     realCt = realCt + 1
   End If
   ct = ct + 1
 Next
 Next
 Next
 
  
 Text1.Text = Text1.Text & crlf & realCt & Str(ct) & "         " & "(" & dblCt & ")"
 Text1.Text = Text1.Text & crlf & realCt / ct & crlf
  
End Sub


  Posted by Charlie on 2016-09-20 15:43:45
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