+0

# math 142

0
140
2

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
Sort:

#1
+18599
+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
$$\frac{4320}{8!} =\frac{4320}{40320} =0.10714285714 =10.7143 \%$$

heureka  Jun 28, 2017
#2
+26231
+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

### 28 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details