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Notice that $Y \ge 3$, hence $P(Y \ge 5)=1-P(Y=3)-P(Y=4)$. There is exactly one way to get $Y=3$, that is $(X_1,X_2,X_3)=(1,1,1)$. Ways to get $Y=4$ are $(2,1,1),(1,2,1),(1,1,2)$. Hence compute $1- …

Hint: Use the property that pdf integrates to $1$ to solve for $k$.

Guide: If assume independence, then we have $A-2P$ is normal distribution with mean $$\mathbb{E}[A-2P]=\mathbb{E}[A]-2\mathbb{E}[P]=180-2(100)$$ and $$Var[A-2P]=Var[A]+4Va[P]=12^2+4(10^2)$$

From the third step to the fourth step, you did not include any intersection term. \begin{align} &P(A) + P( B \cap A') \cup (C \cap A') \\&=P(A)+ P(B \cap A') + P(C \cap A') \color{blue}{- P(B \cap …

Guide: Taxis arrive according to a Poisson Process, hence you should consider Poisson distribution with mean $15\lambda$ since we should be interested in how many taxis arrive in $15$ minutes.

Let $O$ be the event that the room belongs to Old College Let $R$ be the event that a room has a refrigerator. Let $T$ be the event that a room has a TV \begin{align} P(O|R \cup T) &= \frac{P(O)P(R \ …

Guide: Great, you have solved the parameters. Now, you just have to use the formula $$Pr(X=i) = \begin{cases} \binom{6}{i}\left( \frac13 \right)^i \left( \frac23 \right)^{6-i}& i \in \{ 0, \ldots, …

No idea why do you compute expectation of $Y$ but $$\int_0^1 x^2 \, dx = \frac13 $$ The density must have changed. A possible way to find pdf is to first find the CDF then differentiate it to fin …

If $A$ and $B$ are mutually exclusive, then $A \subset B^c$, Hence $P(A \cap B^c) = P(A)$

I doubt there can be a definite answer unless we find the person who wrote it there and asked him or her. I am guessing that $G$ refers to Gaussian. Could $\mu+C(\sigma)$ be a confidence interval? …

\begin{align} P(Z^2 \leq y) &= P(-\sqrt{y} \leq Z \leq \sqrt{y}) \\ &= F_Z(\sqrt{y}) - F_Z(-\sqrt{y}) \end{align} Differentiating $$f_{Z^2}(y) = \frac{1}{2\sqrt{y}}f_Z(\sqrt{y}) + \frac{1}{2\sqrt{y} …

Guide: Note that $\bar{X}-\bar{Y}$ follows a normal distribution of which you already know that mean ($0$) and standard deviation ($\sigma$). \begin{align}Pr(|\bar{X}-\bar{Y}| > 2) &= Pr(\bar{X}-\b …

Guide: Try computing the probability explicitly. $P(X=1)= \frac{1}{20}$ $P(X=2)=\frac{19}{20}\frac{1}{19}$ Hopefully you can spot a pattern.

If a marble is neither red nor blue. It must be either white or orange. Your answer is correct. $$\frac{45}{75}=\frac{3}{5}$$

Guide: From $$P(X \geq 0 ) = 0.6$$ we have $$P\left(\frac{X-\mu}{\sigma} \geq \frac{0-\mu}{\sigma} \right) = 0.6$$ You should be able to obtain another linear equation from above. With two equat …