Rom

you don't give any choices...

total = p + h + s + c + d = 3h + h + 4c + c + 3(3h) = 13h + 5c

here's the first 100 possible totals

18, 23, 28, 31, 33, 36, 38, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56,
57, 58, 59, 61, 62, 63, 64, 66, 67, 69, 70, 71, 72, 74, 75, 76, 77,
79, 80, 82, 83, 84, 85, 87, 88, 89, 90, 92, 93, 95, 96, 97, 98, 100,
101, 102, 103, 105, 106, 108, 109, 110, 111, 113, 114, 115, 116, 118,
119, 121, 122, 123, 124, 126, 127, 128, 129, 131, 132, 134, 135, 136,
137, 139, 140, 141, 142, 144, 145, 147, 149, 150, 152, 154, 155, 157, 
160, 162, 165, 167, 170, 175, 180

this is correct then?

\(P[2]=\dfrac{8}{10}=\dfrac{4}{5}\\ P[4] = \dfrac 1 5\\ E[N] = 8\cdot 2 \cdot \dfrac 4 5 + 2 \cdot 4 \cdot \dfrac 1 5 = \\ \dfrac{64}{5} + \dfrac 8 5 = \dfrac{72}{5}\)

\(E[X] = 2 \cdot \dfrac{8}{n}+4\cdot \dfrac{n-8}{n} = \\ \dfrac{16}{n}-\dfrac{32}{n} + 4 = \\ 4 - \dfrac{16}{n}\)

\(3.5 = 4 - \dfrac {16}{n}\\ \dfrac{16}{n} = \dfrac 1 2\\ n = 32\\ \text{Thus we must add }32-10 = 22 \\ \text{4's to the bag}\)

Thinking about this for a bit it seems to me the only way you can't form a triangle with the 3 segments is if one of those segments has length greater than or equal to half the length of the original segment.

Suppose our original segment is length 1.

We can get a length 1/2 or greater segment in 3 ways

a) both points are greater than 1/2.

b) both points are less than 1/2.

c) the difference between the two points is greater than 1/2

The region satisfying the union of these constraints is as shown below

And it's easy to eyeball that given underlying uniform distributions that the region of interest (beige)

has an area of 3/4 and thus

\(P[\text{triangle cannot be made from 3 segments formed by 2 random points}]=\dfrac 3 4\\ P[\text{triangle can be made}] = 1 - \dfrac 3 4 = \dfrac 1 4\)

Another set of eyes on this would be helpful.

The problem states that the letters can be arranged in any order.

I take that to mean that (w,x,y,z) and (x,y,z,w) and all it's permutations are the same code.

The question was:

A balloon is losing 2 cubic inches of air per hour.

At what rate is it's radius decreasing when the balloon is 3 inches in radius.

\(\left( \begin{array}{cc} \{1,2,3,4,5\} & \{1,2,3,4,6\} \\ \{1,2,3,4,5\} & \{1,2,4,5,6\} \\ \{1,2,3,4,5\} & \{2,3,4,5,6\} \\ \{1,2,3,4,6\} & \{1,2,3,4,5\} \\ \{1,2,3,4,6\} & \{1,2,4,5,6\} \\ \{1,2,3,5,6\} & \{1,2,4,5,6\} \\ \{1,2,4,5,6\} & \{1,2,3,4,5\} \\ \{1,2,4,5,6\} & \{1,2,3,4,6\} \\ \{1,2,4,5,6\} & \{1,2,3,5,6\} \\ \{1,2,4,5,6\} & \{1,2,4,5,6\} \\ \{1,2,4,5,6\} & \{1,3,4,5,6\} \\ \{1,2,4,5,6\} & \{2,3,4,5,6\} \\ \{1,3,4,5,6\} & \{1,2,4,5,6\} \\ \{2,3,4,5,6\} & \{1,2,3,4,5\} \\ \{2,3,4,5,6\} & \{1,2,4,5,6\} \\ \end{array} \right)\)

This isn't quite correct.

You've treated the groups of three as permutations but the rows, columns and diagonals as combinations.

You need to be consistent in how you treat them.

If we went w/permuations we have 8 x 3! =48 different permuations for the rows, columns, diagonals

Then the probability becomes p = (48)/(504) = 2/21

If we use combinations we have 8 rows, columns, diagonals but then we have

9C3 = 84 possible arrangements of 3 darts on the board.

This gets us again a probability of p = 8/84 = 2/21

I'm afraid this isn't correct.

Ignoring duplicates for a moment there are

\(n = \dbinom{4}{1}\dbinom{4}{3} = 16\) different codes that are generated as described

There is however 1 code that is duplicated

(w,x,y,z) can be obtained by either (x)+(w,y,z) or (z) + (w,x,y)

So we end up with a total of 15 distinct codes.