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6

$\vert E(X) - E(X_n) \vert =\vert E(X-X_n) \vert \leq E(|X-X_n|) \to 0$


2

How about trying by contrapositive: If $E[X_n] \not\to E[X]$, then by linearity, $E[X_n-X]\not\to 0$. And your result should follow.


2

The product will not be even only if you get $1$ or $3$ or $5$ every time you roll the die. These are $3$ out of $6$ options, so the probability for that after $n$ attempts is $(\dfrac{3}{6})^n$. So the probability that the product will not be even after $5$ attempts is $(\dfrac{3}{6})^5$. And therefore, the probability that the product will be even is ...


1

Let $Y=f(U)$. You know that $E(Y)=E(f(U))=\frac{1}{b-a}\int_{a}^{b}f(u) du$. You also know (no?) that $\sum f(U_i)/n =\sum Y_i/n \to E(Y)$ in probability (law of large numbers) if the variance of $Y$ is finite (sufficient condition). Now, $Y$ is bounded (because a continous function on a closed interval is bounded). Hence, it's variance is finite, and ...


1

First of all, make sure that the probabilities add up to $1$: $0.1+0.1+0+0.2+0.4+0.2=1$. Good, so the data given to you represents a "complete system". Now, each event is independent, i.e., for every event, if it occurs then all other events do not occur. So the probability that $X \leq Y$ is simply the sum of the probabilities of the events in which $X ...


1

The set of numbers less than 9 is (1, 2, 4, 8) and we need to choose 3 of them, so we have 4 choose 3 which is 4. This uses 3 of the 7 positions available. Good. Then a 9 has to be selected for one of the positions as well, so there are only 3 positions left. Finally, since there are no repetitions the only numbers left are 10, 11, 15, and 17 ...


1

This is correct. The assumption that the actions (vaccination & leaving the city) are independent is crucial: Imagine that the vaccination works only in the city. In that case the probability of avoiding the disease after leaving the city is then 25/100 even if you are vaccinated.


1

You're missing data that you need to calculate the probability. If we knew the probability of one who neither takes the vaccines nor leaves the city contracting the disease, we could answer the question. Suppose the probability of avoiding infection if you take neither precaution is 0/100. Then we could say "Oh, getting the vaccine reduces your chances ...


1

First assume that $X$ is nonnegative. Let $I_n$ be the indicator function for $\bigcup_{k=0}^n A_k$. Then $$ \sum_{k=0}^n E(X;A_k)=\int_{\cup_{k=0}^n A_k} X(\omega)dP(\omega)=\int_\Omega I_n(\omega)X(\omega)dP(\omega). $$ The first equality above uses the disjointness of the $A_k$'s. The sequence $\{I_nX\}_n$ is increasing and converges to $I_AX$ where $I_A$ ...


1

I think you're supposed to assume (perhaps rather strange assumption) that the results of the previous matches can be used to state the probability of outcomes of future matches. So in any given match, $A$ wins with probability $\frac{6}{12}=\frac12$; $B$ wins with probability $\frac13$; and the match ends in a tie with probability $\frac16$. I believe ...


1

Here's a bit of a rigorous argument: recall that if $A \subseteq B$ ($A$ is contained entirely in $B$) are events, then $\mathbb{P}\left(A\right) \leq \mathbb{P}\left(B\right)$. You can easily prove that $A \cap B\cap C \subseteq A \cap C$, and thus, $\mathbb{P}\left(A \cap B \cap C\right) \leq \mathbb{P}\left(A \cap B\right) = 0$. Since $\mathbb{P}$ is a ...


1

Your sample spaces are not correct. For the first one, you have six possible rolls each time, so the size of the sample space is $6^4$, but that is not needed. The chance of getting at least one six in four rolls is one minus the chance of getting no sixes. What is the chance of not getting a six on one roll? Now you need to succeed at that four times in ...


1

Numbers with leading zeros are typically not counted in questions of this kind. The last digit must be one of: $0, 2, 4$. Case 1: If it is $0$ then there are four possibilities for the first digit: $1,2,3,4$. Case 2: If it is not $0$ then there are three possibilities: $1,3$ and the other non-zero even digit. Then there are three possibilities for the ...


1

$P(A \mid B) = \displaystyle\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{b}$. Now, why does $P(A \cap B) \geq a+b-1$? Think of inclusion-exclusion or just draw venn diagrams. Also, $P(A \cap B) = ab$ if $A$ and $B$ are independent, which I don't think is given in the problem...



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