PartialMathematician

Here is the tree for  part a).

0 child              1st child              2nd child

                                                     | -- Boy (27.04%)

                           _ Boy (52%) -------[

Start                   |                          | -- Girl (24.96%)

    X -------------------|                          

                          |                          | -- Boy (24.96%) 

                          |_ Girl (48%) --------[

                                                     |_ Girl (23.04%)

- PM

B = boy, G = girl

a) 
Probability of 2B is (0.52)(0.52) = 0.2704.
Probability of BG is (0.52)(0.48) = 0.2496 
Probability of GB is the same as the probability of BG, so it is 0.2496.
Probability of 2G is (0.48)(0.48) = 0.2304.

b) 
Number of girls born: 

Table for p(x) 
For zero girls and 2 boys, it is 0.2704.
For 1 girl and 1 boy, it is 0.4992.
For 2 girls, it is 0.2304. 

- PM

There are 7 points, so there are 7! = 5040 ways to arrange the points without considering overcounting.

Let's first consider rotations. If we number the 6 beads on the outside as ABCDEF and the center bead as G, the arrangement of ABCDEFG is considered the same as BCDEFGA. Since there are 6 different locations the outer bead A can be, (A cannot be in the center) we divide 5040 by 6 to get 840.

Let's now consider reflections. We can draw six axes for the beads to reflect on, and ABC G(center) DEF is considered the same as DEF G(center) ABC. We divide 840 by 6 to get our answer, \(\boxed{140}\).

Hope this helps, 

- PM 

\(\dfrac{7\cdot(5!)}{6} = \boxed{140}\)

- PM

Question 2) All of the items must be sold to profit at least $1, so we must choose the cheapest item to make sure that the markup percentage suits all of the items. 

The cheapest item is the Spin Top, costing $1.50  (Man, that is an expensive top!). $1 is 2/3, or 66.6%, of the price of the top, so we must increase the markup price by at least \(\boxed{67\%}\) to get at least a $1 profit on all items.

Hope this helps,

- PM 

Question 1) 2 necklaces are $15 without marking up the price, so each necklace is $7.50. If the price is marked up by 15%, we can multiply 15 by \(\dfrac{115}{100} \Rightarrow \dfrac{23}{20}\) to get 17.25.

Each individual necklace is $8.625, or $8.63. Each necklace will be sold for $\(\boxed{8.63}\), and the profit for each necklace is 8.63 - 7.50 = \(\boxed{\$ 1.13}\). 

- PM

94 - 27 = 67, so between the 27th and 94th entries, there are 67-1 = 66 entries. The entries before the 94th entry were probably one greater than the entry after, so entry 93 is 28, entry 92 is 29, etc. This makes sense because the average of E92 and E94 is 28, which is the value in E93. 

If the 27th entry is 94, then the 26th entry must be 95, the 25th entry must be 96, etc. From the 27th entry, there are 26 entries ahead of it, so the finrst entry is 94 + 26 = 120. The first entry is \(\boxed{120}\).

Hope this helps, 

- PM

Two distinct possible values for B.

S(n) = \(\dfrac{Sn + 1}{(n+1)!}\).