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 #6
avatar+26367 
+1

n=1/1+2 + 1/1+2+3 + ...... + 1/1+2+3+....+2014 + 2/2015, n=?

 

\(\begin{array}{|rcll|} \hline n &=& \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4} + \ldots + \frac{1}{1+2+3+\ldots+2013} + \frac{1}{1+2+3+\ldots+2014} + \frac{2}{2015} \\ \hline \end{array} \)

 

 

\(\begin{array}{rcll} n &=& \frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4} + \ldots + \frac{1}{1+2+3+\ldots+2013} + \frac{1}{1+2+3+\ldots+2014} + \frac{2}{2015} \\ &=& \frac{1}{\left(\frac{1+2}{2}\right)\cdot 2} +\frac{1}{\left(\frac{1+3}{2}\right)\cdot 3} +\frac{1}{\left(\frac{1+4}{2}\right)\cdot 4} + \ldots + \frac{1}{\left(\frac{1+2013}{2}\right)\cdot 2013} + \frac{1}{\left(\frac{1+2014}{2}\right)\cdot 2014} + \frac{2}{2015} \\ &=& \frac{2}{2\cdot 3} +\frac{2}{3\cdot 4} +\frac{2}{4\cdot 5} + \ldots + \frac{2}{2013\cdot 2014} + \frac{2}{2014\cdot 2015} + \frac{2}{2015} \\ &=& 2\times \Big( \frac{1}{2015} + \frac{1}{2\cdot 3} +\frac{1}{3\cdot 4} +\frac{1}{4\cdot 5} + \ldots + \frac{1}{2013\cdot 2014} + \frac{1}{2014\cdot 2015} \Big)\\\\ && \mathbf{\frac{1}{a\cdot (a+1)} = \frac{1}{a} - \frac{1}{a+1}} \\\\ &=& 2\times \Big[ \frac{1}{2015} + \left(\frac{1}{2}-\frac{1}{3}\right) +\left(\frac{1}{3}-\frac{1}{4}\right) +\left(\frac{1}{4}-\frac{1}{5}\right) + \ldots \\ && + \left(\frac{1}{2013}-\frac{1}{2014}\right) + \left(\frac{1}{2014}-\frac{1}{2015}\right) \Big] \\\\ &=& 2\times \Big[ \frac{1}{2015} + \frac12+ \underbrace{\left(-\frac{1}{3}+\frac{1}{3}\right)}_{=0} +\underbrace{\left(-\frac{1}{4}+\frac{1}{4}\right) }_{=0} +\underbrace{\left(-\frac{1}{5}+\frac{1}{5}\right)}_{=0} + \ldots \\ && + \underbrace{\left(-\frac{1}{2013}+\frac{1}{2013}\right)}_{=0} + \underbrace{\left(-\frac{1}{2014}+\frac{1}{2014}\right)}_{=0} -\frac{1}{2015} \Big] \\\\ &=& 2\times \left(\frac{1}{2015}+ \frac12 -\frac{1}{2015} \right) \\\\ &=& 2\times \left(\frac12 \right) \\\\ &\mathbf{=}& \mathbf{1} \\ \end{array}\)

 

laugh

06.07.2017
 #19
avatar+26367 
+1

In the Small State Lottery,
three white balls are drawn at random from twenty balls labled 1-20 and
a blue Superball is drawn from ten balls labeled 21-30.
To win a prize you must match at least two of the white balls or match the blue Superball.
If you buy a ticket what is the probability that you would win a prize?

 

Let W be the probability of choosing 2 white balls out of 20 or 3 white balls out of 20 .

Let \(\overline{\mathbf{W}}\)be not W.

 

Let B be the probability of choosing one blue ball out of 10.

Let \(\overline{\mathbf{B}}\) be not B

 

\(\begin{array}{|rcll|} \hline W &=& \dfrac{\binom{3}{2}\binom{17}{1} + \binom{3}{3}\binom{17}{0} }{\binom{20}{3}} = \dfrac{13}{285} \\\\ \overline{W} &=& 1 - W = 1-\dfrac{13}{285} = \dfrac{272}{285} \\\\ B &=& \dfrac{\binom{1}{1}\binom{9}{0} }{\binom{10}{1}} = \dfrac{1}{10} \\\\ \overline{B} &=& 1 - B = 1 - \dfrac{1}{10} = \dfrac{9}{10} \\ \hline \end{array} \)

 

\(\begin{array}{|c|c|c|c|} \hline & \mathbf{B} & \mathbf{\overline{B}} \\ & \dfrac{1}{10} & \dfrac{9}{10} \\ \hline \mathbf{W} & W\cap B & W\cap \overline{B} \\ \dfrac{13}{285} & \dfrac{13}{285} \times \dfrac{1}{10} & \dfrac{13}{285} \times \dfrac{9}{10} \\ \hline \mathbf{\overline{W}} & \overline{W}\cap B & \overline{W}\cap \overline{B} \\ \dfrac{272}{285} & \dfrac{272}{285} \times \dfrac{1}{10} & \dfrac{272}{285} \times \dfrac{9}{10} \\ \hline \end{array} \)

 

The probability of not a prize is
\(\overline{W}\cap \overline{B} = \dfrac{272}{285} \times \dfrac{9}{10} \\\)


The probability of a prize is
\(\begin{array}{|rcll|} \hline && 1- \overline{W}\cap \overline{B} \\ &=& 1- \left(\frac{272}{285} \times \frac{9}{10} \right) \\ &=& \frac{285\cdot 10- 9\cdot 272}{285\cdot 10} \\ &=& \frac{402}{2850} \\ &=& \frac{67\cdot 6}{475\cdot 6} \\ &=& \frac{67}{475} \\ &=& 0.14105263158 \\ \hline \end{array} \)

 

laugh

05.07.2017
 #1
avatar+26367 
+1

If a permutation is chosen at random from the letters "AAABBBCD",

what is the probability that it begins with at least 2 A's?
Round your answer to 6 decimal places as needed.

 

     1. 36 x AAABBBCD
     2. 36 x AAABBBDC
     3. 36 x AAABBCBD
     4. 36 x AAABBCDB
     5. 36 x AAABBDBC
     6. 36 x AAABBDCB
     7. 36 x AAABCBBD
     8. 36 x AAABCBDB
     9. 36 x AAABCDBB
    10. 36 x AAABDBBC
    11. 36 x AAABDBCB
    12. 36 x AAABDCBB
    13. 36 x AAACBBBD
    14. 36 x AAACBBDB
    15. 36 x AAACBDBB
    16. 36 x AAACDBBB
    17. 36 x AAADBBBC
    18. 36 x AAADBBCB
    19. 36 x AAADBCBB
    20. 36 x AAADCBBB
    21. 36 x AABABBCD
    22. 36 x AABABBDC
    23. 36 x AABABCBD
    24. 36 x AABABCDB
    25. 36 x AABABDBC
    26. 36 x AABABDCB
    27. 36 x AABACBBD
    28. 36 x AABACBDB
    29. 36 x AABACDBB
    30. 36 x AABADBBC
    31. 36 x AABADBCB
    32. 36 x AABADCBB
    33. 36 x AABBABCD
    34. 36 x AABBABDC
    35. 36 x AABBACBD
    36. 36 x AABBACDB
    37. 36 x AABBADBC
    38. 36 x AABBADCB
    39. 36 x AABBBACD
    40. 36 x AABBBADC
    41. 36 x AABBBCAD
    42. 36 x AABBBCDA
    43. 36 x AABBBDAC
    44. 36 x AABBBDCA
    45. 36 x AABBCABD
    46. 36 x AABBCADB
    47. 36 x AABBCBAD
    48. 36 x AABBCBDA
    49. 36 x AABBCDAB
    50. 36 x AABBCDBA
    51. 36 x AABBDABC
    52. 36 x AABBDACB
    53. 36 x AABBDBAC
    54. 36 x AABBDBCA
    55. 36 x AABBDCAB
    56. 36 x AABBDCBA
    57. 36 x AABCABBD
    58. 36 x AABCABDB
    59. 36 x AABCADBB
    60. 36 x AABCBABD
    61. 36 x AABCBADB
    62. 36 x AABCBBAD
    63. 36 x AABCBBDA
    64. 36 x AABCBDAB
    65. 36 x AABCBDBA
    66. 36 x AABCDABB
    67. 36 x AABCDBAB
    68. 36 x AABCDBBA
    69. 36 x AABDABBC
    70. 36 x AABDABCB
    71. 36 x AABDACBB
    72. 36 x AABDBABC
    73. 36 x AABDBACB
    74. 36 x AABDBBAC
    75. 36 x AABDBBCA
    76. 36 x AABDBCAB
    77. 36 x AABDBCBA
    78. 36 x AABDCABB
    79. 36 x AABDCBAB
    80. 36 x AABDCBBA
    81. 36 x AACABBBD
    82. 36 x AACABBDB
    83. 36 x AACABDBB
    84. 36 x AACADBBB
    85. 36 x AACBABBD
    86. 36 x AACBABDB
    87. 36 x AACBADBB
    88. 36 x AACBBABD
    89. 36 x AACBBADB
    90. 36 x AACBBBAD
    91. 36 x AACBBBDA
    92. 36 x AACBBDAB
    93. 36 x AACBBDBA
    94. 36 x AACBDABB
    95. 36 x AACBDBAB
    96. 36 x AACBDBBA
    97. 36 x AACDABBB
    98. 36 x AACDBABB
    99. 36 x AACDBBAB
   100. 36 x AACDBBBA
   101. 36 x AADABBBC
   102. 36 x AADABBCB
   103. 36 x AADABCBB
   104. 36 x AADACBBB
   105. 36 x AADBABBC
   106. 36 x AADBABCB
   107. 36 x AADBACBB
   108. 36 x AADBBABC
   109. 36 x AADBBACB
   110. 36 x AADBBBAC
   111. 36 x AADBBBCA
   112. 36 x AADBBCAB
   113. 36 x AADBBCBA
   114. 36 x AADBCABB
   115. 36 x AADBCBAB
   116. 36 x AADBCBBA
   117. 36 x AADCABBB
   118. 36 x AADCBABB
   119. 36 x AADCBBAB
   120. 36 x AADCBBBA
 
      4320 x AA \(\ldots\)


The probability that "AAABBBCD" begins with at least 2 A's is
\(\frac{4320}{8!} =\frac{4320}{40320} =0.10714285714 =10.7143 \% \)

 

laugh

28.06.2017