Consider a string of n quaternary digits, randomly created.
Evaluate the probability that in this string the quantity
of an a priori defined digit, say zero, is even.
Double check your result by
either
a. direct count and comparison (e.g. n=4)
or
b. getting the same result by two distinct methods.
Sigma{i=0 to floor(n/2)} (1/4)^(2i) * (3/4)^(n-2i) * C(n,2i)
n probability
1 .75
2 .625
3 .5625
4 .53125
5 .515625
6 .5078125
7 .50390625
8 .501953125
9 .5009765625
10 .50048828125
11 .500244140625
12 .5001220703125
13 .50006103515625
14 .500030517578125
15 .500015258789063
16 .500007629394531
17 .500003814697266
18 .500001907348633
19 .500000953674316
20 .500000476837158
21 .500000238418579
22 .50000011920929
23 .500000059604645
24 .500000029802322
25 .500000014901161
from
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
For n = 1 To 25
p = 0
For i = 0 To n / 2
p = p + (1 / 4) ^ (2 * i) * (3 / 4) ^ (n - 2 * i) * combi(n, 2 * i)
Next i
Text1.Text = Text1.Text & n & Str(p) & crlf
Next
Text1.Text = Text1.Text & crlf & " done"
End Sub
Function combi(n, r)
c = 1
For i = n To n - r + 1 Step -1
c = c * i
Next
For i = 2 To r
c = c / i
Next
combi = c
End Function
Double check via choice a:
direct count of even number of zeros (evens marked with *):
0000 4 *
0001 3
0002 3
0003 3
0010 3
0011 2 *
0012 2 *
0013 2 *
0020 3
0021 2 *
0022 2 *
0023 2 *
0030 3
0031 2 *
0032 2 *
0033 2 *
0100 3
0101 2 *
0102 2 *
0103 2 *
0110 2 *
0111 1
0112 1
0113 1
0120 2 *
0121 1
0122 1
0123 1
0130 2 *
0131 1
0132 1
0133 1
0200 3
0201 2 *
0202 2 *
0203 2 *
0210 2 *
0211 1
0212 1
0213 1
0220 2 *
0221 1
0222 1
0223 1
0230 2 *
0231 1
0232 1
0233 1
0300 3
0301 2 *
0302 2 *
0303 2 *
0310 2 *
0311 1
0312 1
0313 1
0320 2 *
0321 1
0322 1
0323 1
0330 2 *
0331 1
0332 1
0333 1
1000 3
1001 2 *
1002 2 *
1003 2 *
1010 2 *
1011 1
1012 1
1013 1
1020 2 *
1021 1
1022 1
1023 1
1030 2 *
1031 1
1032 1
1033 1
1100 2 *
1101 1
1102 1
1103 1
1110 1
1111 0 *
1112 0 *
1113 0 *
1120 1
1121 0 *
1122 0 *
1123 0 *
1130 1
1131 0 *
1132 0 *
1133 0 *
1200 2 *
1201 1
1202 1
1203 1
1210 1
1211 0 *
1212 0 *
1213 0 *
1220 1
1221 0 *
1222 0 *
1223 0 *
1230 1
1231 0 *
1232 0 *
1233 0 *
1300 2 *
1301 1
1302 1
1303 1
1310 1
1311 0 *
1312 0 *
1313 0 *
1320 1
1321 0 *
1322 0 *
1323 0 *
1330 1
1331 0 *
1332 0 *
1333 0 *
2000 3
2001 2 *
2002 2 *
2003 2 *
2010 2 *
2011 1
2012 1
2013 1
2020 2 *
2021 1
2022 1
2023 1
2030 2 *
2031 1
2032 1
2033 1
2100 2 *
2101 1
2102 1
2103 1
2110 1
2111 0 *
2112 0 *
2113 0 *
2120 1
2121 0 *
2122 0 *
2123 0 *
2130 1
2131 0 *
2132 0 *
2133 0 *
2200 2 *
2201 1
2202 1
2203 1
2210 1
2211 0 *
2212 0 *
2213 0 *
2220 1
2221 0 *
2222 0 *
2223 0 *
2230 1
2231 0 *
2232 0 *
2233 0 *
2300 2 *
2301 1
2302 1
2303 1
2310 1
2311 0 *
2312 0 *
2313 0 *
2320 1
2321 0 *
2322 0 *
2323 0 *
2330 1
2331 0 *
2332 0 *
2333 0 *
3000 3
3001 2 *
3002 2 *
3003 2 *
3010 2 *
3011 1
3012 1
3013 1
3020 2 *
3021 1
3022 1
3023 1
3030 2 *
3031 1
3032 1
3033 1
3100 2 *
3101 1
3102 1
3103 1
3110 1
3111 0 *
3112 0 *
3113 0 *
3120 1
3121 0 *
3122 0 *
3123 0 *
3130 1
3131 0 *
3132 0 *
3133 0 *
3200 2 *
3201 1
3202 1
3203 1
3210 1
3211 0 *
3212 0 *
3213 0 *
3220 1
3221 0 *
3222 0 *
3223 0 *
3230 1
3231 0 *
3232 0 *
3233 0 *
3300 2 *
3301 1
3302 1
3303 1
3310 1
3311 0 *
3312 0 *
3313 0 *
3320 1
3321 0 *
3322 0 *
3323 0 *
3330 1
3331 0 *
3332 0 *
3333 0 *
asterisked lines show an even number of zeros
136 out of 256 = .53125
from
For a = 0 To 3
For b = 0 To 3
For c = 0 To 3
For d = 0 To 3
Text1.Text = Text1.Text & a & b & c & d & " "
subct = 0
If a = 0 Then subct = subct + 1
If b = 0 Then subct = subct + 1
If c = 0 Then subct = subct + 1
If d = 0 Then subct = subct + 1
Text1.Text = Text1.Text & subct
If subct Mod 2 = 0 Then Text1.Text = Text1.Text & " *": ct = ct + 1
Text1.Text = Text1.Text & crlf
Next
Next
Next
Next
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Posted by Charlie
on 2017-08-09 11:34:37 |
solution (spoiler)
