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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.

Guest Jun 28, 2017
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2+0 Answers

 #1
avatar+18777 
+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

heureka  Jun 28, 2017
 #2
avatar+26366 
+1

Or simply:  

 

probability that first letter is an A:  p1 = 3/8

given that the first letter is an A the probability that the second letter is also an A: p2 = 2/7

 

Overall probability = p1*p2 = (3/8)*(2/7) → 0.107143  

.

Alan  Jun 28, 2017

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