Rom

I'm seeing that 

P[win at home] = 0.2 = 0.8 x 0.25 = P[win] x P[at home]

therefore the two are independent.

1) 1 and 2 are binomial distributions

3) 1 and 2 are likely to be approximately normally distributed.

Only approximately though as neither ages nor heights can be negative.

The normal distribution, regardless of the mean and variance, has non-zero probability of negative numbers.

\(\text{Find }0 \leq n < 18 \ni n \equiv -11213141 \pmod{18}\)

\(\dfrac{11213141}{18} = 622952 + \dfrac{5}{18} \text{ so }\\ -11213141 + 622953\cdot 18 = 13 \equiv -11213141 \pmod{18}\)

\(\text{If rain and lightning are independent then }\\ P[\text{rain AND lightning}]=P[\text{rain}]P[\text{lightning}]\\ 0.15 \neq 0.6 \cdot 0.3 = 0.18\\ \text{that's choice C}\)

\(P[\text{rain AND lightning}] = \\ P[\text{lightning |raining}]P[\text{raining}] = \\ (0.1)(0.6) = 0.06 = 6\%\\ \text{Choice B}\)

that's some interesting terminology...

A line "clamped" by v?

Does that mean the line passes through v?

10/11 is correct

14/22 + 10/22 - 4/22 = 20/22 = 10/11

You omitted the case where all the questions were answered true, i.e. the C(10,0) case.

That brings the number to 1024 - 1 - 10 - 45 = 968

\(1)~\text{answered in other thread}\)

\(2)~\dfrac{2}{10}\dfrac 1 2 = \dfrac{1}{10}\)

\(3)~\dfrac{\dbinom{2}{1}\dbinom{3}{1}}{\dbinom{10}{2}} = \dfrac{6}{45}=\dfrac{2}{15}\)

\(4)~\dfrac 1 2 \dfrac 1 6 = \dfrac{1}{12}\)

\(5)~\dfrac{10}{18}\dfrac{8}{17} = \dfrac{40}{153}\)

1/2 x 10^-6m = 500 x 10^-9m = 500 nm is the correct answer

\(1)~ \dfrac{26}{52}\dfrac{13}{51} = \dfrac{13}{102}\)

\(2) ~\dfrac{3}{30}\dfrac{2}{29} = \dfrac{1}{145}\)

\(3)~\dfrac{26}{52}\dfrac{13}{52}=\dfrac 1 8\)

\(4)~\dfrac 2 6 \dfrac 1 6 = \dfrac {1}{18}\)