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# math 142

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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?

Guest Jun 28, 2017
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#1
+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?

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
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
33. 36 x AABBABCD
34. 36 x AABBABDC
35. 36 x AABBACBD
36. 36 x AABBACDB
39. 36 x AABBBACD
42. 36 x AABBBCDA
43. 36 x AABBBDAC
44. 36 x AABBBDCA
45. 36 x AABBCABD
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
60. 36 x AABCBABD
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
85. 36 x AACBABBD
86. 36 x AACBABDB
88. 36 x AACBBABD
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

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
+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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