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Ten 6-sided dice are rolled. What is the probability that exactly three of the dice show a 1?

Express your answer as a decimal rounded to the nearest thousandth. 

thanks!

\(\begin{array}{|rcll|} \hline && \binom{10}{3} \times \left(\frac16 \right)^3 \times \left(\frac56 \right)^7 \\\\ &=& \frac{120 \times 5^7 }{6^{10}} \\ \\ &=& \frac{9375000}{60466176} \\ \\ &=& 0.15504535957 \ (15.5 \% ) \\ \hline \end{array}\)

\(\begin{array}{lcll} \text{possibilities: 10 dices, three "1" } &=& \binom{10}{3} \\ \text{probability: 3dices, three "1" } &=& \left(\frac16 \right)^3 \\ \text{probability: 7dices, no "1" } &=& \left(\frac56 \right)^7 \\ \end{array}\)

Find

(a) the number of subsets and

(b) the number of proper subsets of the set.

{x|x is an odd integer between -8 and 8}

Let A = { -7, -5, -3, -1, 1, 3, 5, 7 }

Number of elements = 8

(a)  A has \(2^8 = \mathbf{256} \)  subsets

(b)  A has \(2^8-1 = \mathbf{255}\)  proper subsets

Let A = {5, 6,7,8,9,10}

A. How many subsets does A have?

B. How many proper subsets does A have?

A. A has \(2^6 = \mathbf{64} \)  subsets
B. A has \(2^6-1 = \mathbf{63}\)  proper subsets

Let A = {4, 5, 6, 7, 8, 10, 11}

A. How many subsets does A have?

B. How many proper subsets does A have?

A. A has \(2^7 = \mathbf{128}\)  subsets

B. A has \(2^7 -1 = \mathbf{127}\) proper subsets

The digits 1,2,3,4 and 5 can be arranged to form many different 5-digit positive integers with five distinct digits.
 In how many such integers is the digit 1 to the left of the digit 2?

\(\begin{array}{|r|r|r|r|r|r|} \hline & & 1 \text{ left } 2& 1 \text{ left } 2& 1 \text{ left } 2& 1 \text{ left } 2 \\ \hline n & \text{permutation} & \text{ distance }0 & \text{ distance }1 & \text{ distance }2 & \text{ distance }3 \\ \hline 1. & 12345& 0 \\ 2. & 12354& 0 \\ 3. & 12435& 0 \\ 4. & 12453& 0 \\ 5. & 12543& 0 \\ 6. & 12534& 0 \\ 7. & 13245 && 1 \\ 8. & 13254 && 1 \\ 9. & 13425 &&& 2 \\ 10. & 13452 &&&& 3 \\ 11. & 13542 &&&& 3\\ 12. & 13524 &&& 2 \\ 13. & 14325 &&& 2 \\ 14. & 14352 &&&& 3 \\ 15. & 14235 && 1 \\ 16. & 14253 && 1 \\ 17. & 14523 &&& 2 \\ 18. & 14532 &&&& 3 \\ 19. & 15342 &&&& 3 \\ 20. & 15324 &&& 2 \\ 21. & 15432 &&&& 3 \\ 22. & 15423 &&& 2 \\ 23. & 15243 && 1 \\ 24. & 15234 && 1 \\ 25. & 21345 \\ 26. & 21354 \\ 27. & 21435 \\ 28. & 21453 \\ 29. & 21543 \\ 30. & 21534 \\ 31. & 23145 \\ 32. & 23154 \\ 33. & 23415 \\ 34. & 23451 \\ 35. & 23541 \\ 36. & 23514 \\ 37. & 24315 \\ 38. & 24351 \\ 39. & 24135 \\ 40. & 24153 \\ 41. & 24513 \\ 42. & 24531 \\ 43. & 25341 \\ 44. & 25314 \\ 45. & 25431 \\ 46. & 25413 \\ 47. & 25143 \\ 48. & 25134 \\ 49. & 32145 \\ 50. & 32154 \\ 51. & 32415 \\ 52. & 32451 \\ 53. & 32541 \\ 54. & 32514 \\ 55. & 31245 &0 \\ 56. & 31254 &0 \\ 57. & 31425 && 1 \\ 58. & 31452 &&& 2 \\ 59. & 31542 &&& 2 \\ 60. & 31524 && 1 \\ 61. & 34125 &0 \\ 62. & 34152 && 1 \\ 63. & 34215 \\ 64. & 34251 \\ 65. & 34521 \\ 66. & 34512 &0 \\ 67. & 35142 && 1 \\ 68. & 35124 &0 \\ 69. & 35412 &0 \\ 70. & 35421 \\ 71. & 35241 \\ 72. & 35214 \\ 73. & 42315 \\ 74. & 42351 \\ 75. & 42135 \\ 76. & 42153 \\ 77. & 42513 \\ 78. & 42531 \\ 79. & 43215 \\ 80. & 43251 \\ 81. & 43125 &0 \\ 82. & 43152 && 1 \\ 83. & 43512 &0 \\ 84. & 43521 \\ 85. & 41325 && 1 \\ 86. & 41352 &&& 2 \\ 87. & 41235 &0 \\ 88. & 41253 &0 \\ 89. & 41523 && 1 \\ 90. & 41532 &&& 2 \\ 91. & 45312 &0 \\ 92. & 45321 \\ 93. & 45132 && 1 \\ 94. & 45123 &0 \\ 95. & 45213 \\ 96. & 45231 \\ 97. & 52341 \\ 98. & 52314 \\ 99. & 52431 \\ 100. & 52413 \\ 101. & 52143 \\ 102. & 52134 \\ 103. & 53241 \\ 104. & 53214 \\ 105. & 53421 \\ 106. & 53412 &0 \\ 107. & 53142 && 1 \\ 108. & 53124 &0 \\ 109. & 54321 \\ 110. & 54312 &0 \\ 111. & 54231 \\ 112. & 54213 \\ 113. & 54123 &0 \\ 114. & 54132 && 1 \\ 115. & 51342 &&& 2 \\ 116. & 51324 && 1 \\ 117. & 51432 &&& 2 \\ 118. & 51423 && 1 \\ 119. & 51243 &0 \\ 120. & 51234 &0 \\ \hline sum & & 24 & 18& 12 & 6 \\ \hline \end{array}\)

Use the formula S​ = n2 to find the sum of

1​ + 3​ + 5​ + ...​ + 861 = ?

​(Hint: To find​ n, add 1 to the last term and divide by​ 2.)

\(\begin{array}{|rcll|} \hline n &=& \frac{861+1}{2} \\ &=& \frac{862}{2} \\ &=& 431 \\ \hline \end{array} \)

\(\begin{array}{rcll} sum &=& n^2 \\ &=& 431^2 \\ &=& 185761 \\ \end{array}\)

\(\begin{array}{rcll} 1\ + 3\ + 5\ + 7\ + \ ... \ + 859\ + 861 &=& 185761 \\ \end{array}\)

Assume that the student has a cup with 12 writing implements: 6 pencils, 4 ball point pens, and 2 felt-tip pens. In how many ways can the selection be made if no more than one ball point pen is selected if they must choose 4 implements?

\(\begin{array}{|rcll|} \hline && \binom{4}{1}\binom{8}{3} + \binom{4}{0}\binom{8}{4} \\ &=& 4\binom{8}{3} + 1\binom{8}{4} \\ &=& 4\cdot \frac{8}{3}\cdot \frac{7}{2}\cdot \frac{6}{1} + \frac{8}{4}\cdot \frac{7}{3}\cdot \frac{6}{2}\cdot \frac{5}{1} \\ &=& 4\cdot 8 \cdot 7 + 7 \cdot 2 \cdot 5 \\ &=& 224 + 70 \\ &=& 294 \\ \hline \end{array}\)

Help! I don't know how to solve these 2 questions. (Algebra II)

3.

\(\small{ \begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 5} && 4 && -1 && -10 && -23 && -40 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = -1} && -5 && -9 && -13 && -17 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = -4} && -4 && -4 && -4 && \cdots \\ \end{array} }\)

\(\begin{array}{rcl} y &=& \binom{x}{0}\cdot {\color{red}d_0 } + \binom{x}{1}\cdot {\color{red}d_1 } + \binom{x}{2}\cdot {\color{red}d_2 } \\ \end{array}\)

\( \begin{array}{|rcl|} \hline y &=& \binom{x}{0}\cdot {\color{red} 5 } + \binom{x}{1}\cdot {\color{red} (-1) } + \binom{x}{2}\cdot {\color{red} (-4) } \\ &=& 5 - x - \frac{x}{2}\cdot \frac{x-1}{1} \cdot 4 \\ &=& 5 - x - 2x(x-1) \\ &=& 5 - x - 2x^2+2x \\ \mathbf{y} & \mathbf{=} & \mathbf{5 + x - 2x^2} \\ \hline \end{array}\)

4.

\(\small{ \begin{array}{lrrrrrrrrrrrrrrrrr} & {\color{red}d_0 = 20} && 4 && 0 && 20 && 76 && 180 && \cdots \\ \text{1. Difference } && {\color{red}d_1 = -16} && -4 && 20 && 56 && 104 && \cdots \\ \text{2. Difference } &&& {\color{red}d_2 = 12} && 24 && 36 && 48 && \cdots \\ \text{2. Difference } &&&& {\color{red}d_3 = 12} && 12 && 12 && \cdots \\ \end{array} }\)

\(\begin{array}{rcl} y &=& \binom{x}{0}\cdot {\color{red}d_0 } + \binom{x}{1}\cdot {\color{red}d_1 } + \binom{x}{2}\cdot {\color{red}d_2 } + \binom{x}{3}\cdot {\color{red}d_3 } \\ \end{array}\)

\(\begin{array}{|rcl|} \hline y &=& \binom{x}{0}\cdot {\color{red} 20 } + \binom{x}{1}\cdot {\color{red} (-16) } + \binom{x}{2}\cdot {\color{red} 12 } + \binom{x}{3}\cdot {\color{red} 12 } \\ &=& 20 - 16x + \frac{x}{2}\cdot \frac{x-1}{1} \cdot 12 + \frac{x}{3}\cdot \frac{x-1}{2}\cdot \frac{x-2}{1} \cdot 12 \\ &=& 20 - 16x + 6x(x-1) + 2x(x-1)(x-2) \\ &=& 20 - 16x + 6x^2-6x + 2x \left(x^2-3x+2 \right) \\ &=& 20 - 16x + 6x^2-6x + 2x^3-6x^2+4x \\ &=& 20 - 16x -6x + 2x^3 +4x \\ \mathbf{y} & \mathbf{=} & \mathbf{20 -18x +2x^3} \\ \hline \end{array}\)

Use differences to find a pattern in the sequence
4,4,9,20,38,64,99, ...
Assuming that the pattern? continues,
the eighth term should be?

\(\small{ \begin{array}{lrrrrrrrrrrrrrrrrr} & 4 && 4 && 9 && 20 && 38 && 64 && 99 && {\color{red} 144} && \cdots \\ \text{1. Difference } && 0 && 5 && 11 && 18 && 26 && 35 && {\color{red} 45} && \cdots \\ \text{2. Difference } &&& 5 && 6 && 7 && 8 && 9 && {\color{red} 10} && \cdots \\ \text{3. Difference } &&&& 1 && 1 && 1 && 1 && {\color{red} 1} && \cdots \\ \end{array} } \)

\(\begin{array}{|rcrcr|} \hline 9 &+& {\color{red}1} &=& {\color{red}10} \\ 35&+&{\color{red}10} &=& {\color{red}45} \\ 99&+&{\color{red}45} &=& {\color{red}144} \\ \hline \end{array} \)

The eighth term is 99 + 35 + 9 + 1 = 144

Two distinct points $A$ and $B$ are on a circle with center at $O$, and point $P$ is outside the circle such that $\overline{PA}$ and $\overline{PB}$ are tangent to the circle. Find $AB$ if $PA = 12$ and the radius of the circle is 9.