 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Real roots from three dice (Posted on 2016-09-20) 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 No Rating Comments: ( Back to comment list | You must be logged in to post comments.) 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.

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 Please log in:

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