Melody

Ignore the nulls so that P(R) = P(B) = 1/2.

Calculate the chances of you losing 59 times in a row and Compare these chances to the odds of winning the NJ lotto (say 1/150M). risk. Yes, you calculated the numbers correctly and you have an edge. But there are other issues to consider before deciding whether to play. The most significant one is “gambler’s ruin”: eveb though you have the edge and would surely win if you can play forever (Strong Law of Large Numbers), you can easily go broke if you have insufficient capital.

Chance of losing 59 times in a row is   1/(2^59) = approx   1/ 5.76*10^17

1/(150M)   I will assume M means million.    is approx   1/1.5^10^8

You are much much much more likely to win with lotto.  Stick with lotto.  This place is rigged!

\(\int \frac{sin^2\theta \:cos\theta }{\sqrt{1+sin\theta }}\;d\theta\)

let   \(u^2=1+sin\theta\)

If you need more help then show us what you have managed to do on your own. 

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Ps.   I have done this almost to the end. Unless I have made some really silly mistake/s, it appears to work fine.

So if this is not answered 'soon'  you will no longer be interested in the answer?

So you do not want to understand you just want the answer?  ok, Got it.

Understanding takes time. 

When you research and work it out for yourself the understanding will usually be deeper.

Of course, if you just want the answer, that is called cheating.

Why is it urgent?

Sounds to me like your teacher really wants you to think hard and do some research on your own.

Can you say it in words?

For how many positive integers n less than 100 is a multiple of 6?

Lets simplify the expression (mod6)

\(5^n\mod6\\ \equiv(-1)^n\mod6\\ \)

which is -1 when n is odd and +1 when n is even

\(13^{(n+2)}\equiv 1^{(n+2)}\equiv1\mod6\)

\(8^{(n+1)}\equiv 2^{(n+1)}\equiv2*2^n\mod6\\~\\ 14^{(n+3)}\equiv 2^{(n+3)}\equiv8*2^n\equiv2*2^n\mod6\\~\\ 8^{(n+1)}+14^{(n+3)}=4*2^n\mod6\\~\\ \)

so

\(5^n + 8^{(n + 1)} + 13^{(n + 2)} + 14^{(n + 3)}\mod6 \\ \equiv 1+ (-1)^n+4*2^n \mod6\\~\\ \text{When n is odd}\equiv 4*2^n \mod6\\~\\ \text{When n is even}\equiv 2+4*2^n \mod6\\~\\ \)

consider 4*2^n

So that expression is a multiple of 6 for all even values of n.  So that is 49 values,  2 to 98 inclusive

5*25*26

I assume you want to expand this?

(-x^2)*(x^3+3/2x+1)

\( (-x^2)*(x^3+\frac{3}{2}x+1)\)

Muliply each term in the second bracket by -x^2

\(=(-x^2)*(x^3)\quad+\quad (-x^2)* (\frac{3}{2}x)\quad +\quad (-x^2)* (1)\)

now simplify each term separately.