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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.

     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
\(\frac{4320}{8!} =\frac{4320}{40320} =0.10714285714 =10.7143 \% \)

This can be written as:  4log_a2 = log_ax

which becomes;  log_a2⁴ = log_ax  so x = 2⁴ or x = 16

.

Looking to match the general expression   ax^2 + bx + c

1). 2s(s-4) → 2s^2 - 8s :   a = 2,  b = -8,  c = 0.  In this case we have s standing in for x.

2).  (w + 7)(w + 3) → w^2 + 10w + 21:   a = 1,  b = 10,  c = 21.  In this case we have w standing in for x.

Try the rest yourself.  I assume 3) should be (2t -1)(t + 5) or it won't be a quadratic!

.

Looks like x/2 is subtracted at each stage:

2x,  3x/2,  x,  x/2,  0,  -x/2,  -x,  -3x/2,  -2x,  -5x/2, ...

Well , Mr.BB, as I said. I could explain it, but you need to be smarter than the machine to understand it. So, that leaves you out.  I generally subscribe to the philosophy of not teaching pigs to sing because it is a waste of my time and it annoys the pigs.  It is easy to tell by your post that you are annoyed, so both of those points are in play, here.

Anyway, just so you know that I’m not exhaling blarney, I will share the method with someone who can understand it, and then comment on the validity of the calculator’s output. 

Hummm. Let me think, whom I might choose . . . .  

The total  possible arrangements  =  12!     

Tne math books can be arranged in 3! ways

Tne history books can be arranged in 5! ways

Tne science books can be arranged in 4! ways

And these groups can be arranged in 3!  ways

So......the probability that  each category can be kept together is

[ 3! * 5! * 4! * 3! ]  / 12!   ≈  0.000216

We want choose any  5 of the 13 females  and any 5 of the 12 males =

C(13,5) * C(12,5)  = 1019304    (1)

And the total committees possible  =   C(25, 10) = 3268760    (2)

So the probability  =  (1)  / (2)  =   1019304 / 3268760  ≈  0.311832

Problems with very big numbers, such as this question, can easily be solved using logs:

log(365!) =778.399745153...........(1)

log(335!) + log(365^30)=

702.063079291 + 76.8687859338

=778.931865225.............................(2). We just subtract (2) from (1)

778.399745153 - 778.931865225 =-0.532120072 - This is the exponent of 10:

10^-.532120072 =0.293683757........., Which is the same answer.

I only show up occasionally. Maybe once or twice a week for about an hour or so. You are very bright young lady, and I know if you are going to answer a question, there isn't much that I can add to it.

There's no need to be sorry Guest!! It can be helpful to have more than one answer, in case the questioner doesn't understand one of them!

You seem really nice  ...Is there a reason why you haven't gotten an account on here? But...I can understand just wanting to be anonymous too.

Sorry, hectictar! I didn't know you were answering it at the same I was! Otherwise, I wouldn't have touched it.

Let the number of Dimes =D   P.S. This is "slang" for Canadian coin currency!!.

Quarters =1/7D

Loonies =3/10D

Toonies =1/10D

0.10D + 0.25(D/7) + 3D/10 + 2/10D =$44.50

0.10D + 1/28D + 0.3D + 0.2D =$44.50

0.6357142857...D =$44.50    divide both sides by 0.6357142857

D = 70 Dimes

70/7 =10 Quarters

0.3 x 70 =21 Loonies

70 x 1/10 =7 Toonies. So that:

[70x.10] + [10 x 0.25] + [21 x $1] + [7 x $2]=$44.50

This is an interesting question!!

They told us all of the labels are wrong.

That means that the box labeled  " apples and oranges "  does not have both apples and oranges. Instead, it must be all apples or all oranges.

So ask for a sample from the box labeled  " apples and oranges ."  If you get an orange, then the box must be all oranges. (If you get an apple, the box is all apples.)

So...let's swap the label on the box that was originally labeled  " apples and oranges "  with its correct label that you know now.

That leaves one label untouched. We know that it is incorrect, since all the labels were wrong to begin with. So, swap this label with the  " apples and oranges "  label.

Now they are all correct! And we only needed one sample. 

Only one sample is sufficient!. Take a sample from the box labeled "apples and oranges". Whatever the sample is, that is right label for that box, since they are all labeled wrong. For example, if the sample was an apple, then that box contains only apples. The other two boxes are labeled "apples" and " oranges". Now we go to the box labeled "oranges". Since we have labeled the first box correctly as "apples", then that box must be "apples and oranges". And the remaining box labeled "apples" must be "oranges".

By the way, Ms. Big Mouth, Alan did exactly as I did to arrive at exactly the same answer, since it is the same question!! If you are too STUPID to understand that, I can't help you!.

60 inches. I suggest you Start chopping wood.

96 inches

Ok, Ms. Big Mouth! Why don't you show us how?? calculate 365!, step by step. Just maybe will learn something from you for a change, instead of just shooting off your goddamned Big Mouth! Go ahead.

Well, Mr. Blarney Bag, like usual, you are wrong. This calculator can display a fully resolved mantissa of at least 2048 digits. You just have to know how to use it. I could explain it but you need to be smarter than the machine to understand it. So, that leaves you out.

This equation is a solution to the probability of two persons NOT sharing the same birthday (date) in a group of thirty persons.

Also......1/9   =  1 / (3²)

And.....x^(2 - y) is what you get if you replace  " y "  with  " y - 1 "  in:     \(x^{1-y}\)

\(x^{1-(y-1)}=x^{1-y+1}=x^{2-y}\)          

$420 / (2 1/2) =$420 x 2/5 =$168 Simple interest for 1 Year

$168 / $1,200 =0.14 x 100 = 14% - Simple annual interest  rate.

100/5 = 20. So 100 is the "dividend", 5 is the "divisor" and 20 is the "quotient".

sum of two cubes:

a³ + b³   =   (a + b)(a² - ab + b²)

difference of two cubes:

a³  - b³   =   (a - b)(a² + ab + b²)