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4

This seems to fall easily to linearity of expectation. Let $a_1,a_2\dots a_n$ be your sequence of outcomes. Let $X_1$ be the indicator variable that $a_1,a_2,a_3,a_4=H$ and $a_5=T$ Let $X_{97}$ be the indicator variable that $a_{96}=T$ and $a_{97},a_{98},a_{99},a_{100}=H$ Finally, for $2\leq n \leq 96$ let $X_n$ be the indicator that $a_{n-1}=T,a_n,a_{n+1}... 3 For$i=1$to$97$, let random variable$X_i$be equal to$1$if there is a string of HHHH that begins at$i$and does not extend, and let$X_i=0$otherwise. We want the expectation of$X_1+\cdots+X_{97}$. By the linearity of expectation, this is$E(X_1)+\cdots+E(X_{97})$. So we need only calculate the$\Pr(X_i)=1$. These are not all equal. If$i=1$or$...

2

In any sequence of $100$ flips, “mark” each spot $HHHH$ occurs (and occurs not within a longer run of $H$’s) between the middle $H$’s. For example, the sequence $$HHTHHHTTTHHHHTHT\dots TTHHHHHTHTHHTTHHHH$$ would be marked $${HHTHHHTTTHH}^{\color{red}|\!}{HHTHT\dots TTHHHHHTHTHHTTHH}^{\color{red}|\!}{HH}^{}.$$ Among the $97$ positions where marks could occur ...

2

To answer your first question: The probability that ten six-sided dice land with the same number is equal to the probability that nine of the six-sided dice have the same number as the tenth i.e. $1/6^9$. Your second question does not make sense, unless you are asking how many rolls you need to make for the probability of ten six sided dice landing on the ...

2

I am sure this will not be useful to you, but it is generally not true. Let $X_1$ take the values $\{ 1, 10 \}$ with equal probability, and $X_2$ take values $\{1,9\}$ with equal probability. Then what are $E[X_1 \; | \; X_1 + X_2 = 11]$ and $E[X_2 \; | \; X_1 + X_2 = 10]$?

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If $X\sim \chi^2_n$ then $\Pr(X\ge 0) =1$. But $\Pr(X_1^2+X_2^2 - X_3^2 \ge 0 ) <1$, because the probability that $X_1^2+X_2^2$ is small and $X_3^2$ is large is not $0$.

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Shouldn't the sum at the denominator be replaced with an integral ? Up to a normalizing constant, your target distribution is : $$p(x) \propto \exp \big( -(x^{2}-\mu^{2})^2 \big).$$ Its normalized version would be : $$\tilde{p}(x) = Cp(x) \quad \text{with} \quad \frac{1}{C}=\int_{\mathbb{R}} \exp\big( -(t^2 - \mu^2)^2 \big) \, dt.$$ Given my humble ...

2

The answer is clearly player $A$. Once the first run of three appears, $B$ can only win if the run continues (assuming it can), but $A$ can win from every subsequent card based on one or usually two possible directions for the run. This is true even without your addition of pivot cards, which make the odds even better for $A$, since then $A$ can win in an ...

1

In the second problem, it is easier to use memorylessness. The distribution of $X-1$, given that $X\gt 1$, is the same as the unconditional distribution of $X^3$. Thus our conditional expectation is $\int_0^\infty x^3e^{-x}\,dx$. In the first problem, I would prefer to say the distribution of $Y$, given that $Y\gt 3$, is the same as the distribution of $3+Y^... 1 Let$Z=(X-\mu)/2$so that$Y=Z^2$. Then for$t<1/2$, the substitution$u=\sqrt{2(1/2-t)} z$gives $$M_Y(t) = E[e^{tZ^2}] = \frac{1}{\sqrt{2\pi}} \int e^{(t-1/2)z^2} \mathop{dz} = \frac{1}{\sqrt{2(1/2-t)}}\frac{1}{\sqrt{2\pi}}\int e^{-u^2/2} = \frac{1}{\sqrt{1-2t}}.$$ 1 Let A be the set that have auto insurance and B the set that have homeowners insurance and U denote universal set. You want to maximize, $$P(B^c|A)$$ $$P(B^c|A)= \frac{P(B^c \cap A)}{P(A)}$$ $$P(B^c|A)= \frac{P(A\setminus A\cap B)}{P(A)}$$ $$P(B^c|A)= \frac{P(A) - P(A\cap B)}{P(A)}$$ $$P(B^c|A) \leq 1-\frac{0.15}{0.75}$$ $$P(B^c|A) \leq 0.8$$ 1 With$p:=P(X\le 1)$, we have$2=\Bbb E[X]\le p\cdot 1+(1-p)\cdot 10$, hence$p\le \frac 89$. 1 Because $$E(X) \le 1\cdot 8/9 + 10\cdot 1/9 = 18/9 = 2$$ If the probability is higher than$8/9$then the$E(X)$must be smaller than$2$, contradiction 1 The odds are$1$to$6^9-1$. The probability is$\frac1{6^9}$. This is because there are$6$choices for the face and$6^{10}$values for$10$dice. After$n$throws, the chances of getting all$10$the same at least once is $$1-\left(\frac{6^9-1}{6^9}\right)^n$$ It never is exactly$1$. However, to get a$50\%$chance of seeing all faces the same, you ... 1 If there are$k$lines not in use, then it is the same as$6-k$lines being in use. If between$2$and$4$lines, inclusive, are not in use, then that is the same as between$2$and$4$lines, inclusive, being in use. ($2$lines not being in use is the same as$4$lines being in use, etc.) If at least$4$lines are not in use, then this is the same as at ... 1 Fifth problem:$2$lines are not in use if and only if$4$are in use.$3$lines are not in use if and only if$3$are in use.$4$lines are not in use if and only if$2$are in use. So we need to add the probabilities that$4$,$3$,$2\$ are in use.

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