# Tag Info

### What can I say about $P(B|A_1 \cap A_2)$ if I know that $P(B|A_1), P(B|A_2) \approx 1$?

The event $A_1\cap A_2$ may be much smaller than either of the events $A_1$ or $A_2$. So even if most outcomes of the events are also outcomes of event $B$, that does not require that any outcomes of ...

### Computing conditional expectation from a given random variable

\begin{gather*} P_{X/\mathcal{G}}(A)=\frac{P((X\in A)\cap\{a\})}{P(\{a\})}I_{\{a\}}(X)+\frac{P((X\in A)\cap\{b,c\})}{P(\{b,c\})}I_{\{b,c\}}(X)\\ E[X/\mathcal{G}](\omega)=\sum_{i=1}^3iP_{X/\mathcal{G}}(...
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1 vote

### What can I say about $P(B|A_1 \cap A_2)$ if I know that $P(B|A_1), P(B|A_2) \approx 1$?

The general principle that you are suggesting is not true. Here is an example. Suppose I am the greatest chess player that ever lived. I have two students. They will be attending a conference with 998 ...
• 3,308
1 vote
Accepted

### Computing conditional expectation from a given random variable

The conditional expectation must be measurable with respect to $\mathcal F,$ therefore the inverse image of $Y(b)$ must belong to $\mathcal F = \{\emptyset,\{a\},\{b,c\},\Omega\}.$ Since the inverse ...
• 850
1 vote
Accepted

• 71.1k

### Can these problems be solved using the same technique?

For the second problem $P(\operatorname{desired})=\frac{4}{52}×\frac{4}{51}×\frac{4}{50}×6×\frac{1}{49}$

### Probability of $r$ consecutive heads occurring exactly once in infinite tosses

To answer the first question being asked in the post: no, the reformulation is not equivalent—in part because the original question asks for a probability, while the reformulation sets up a game but ...
• 80k
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• 169k

### Ordered Chain Rule of Probability

The issue in the tree is that when you write $P(A,B)$ what you should be writing is $P(A_1,B_2)$ - if the sampling is done without replacement, $P(A)$ can be different at each set of branches...
• 2,593
Accepted

### How to determine which event is the conditional event?

It's neither. For conditional probability you assume that one event happened, and then ask for the probability that an additional event also happened. Here, you aren't assuming either event. Here ...
• 71.1k
1 vote
Accepted

### Question 29 from Chapter 3 of A first course to probability from Sheldon Ross ed. 10

What you have calculated for first strategy and your assumption are both correct but there is an error in your answer for strategy $2$. In $2$nd strategy for a $3$ occupant house member to be chosen ...
1 vote
Accepted

### Question 11 from chapter 3 of Sheldon Ross book A First course to probability 10th ed.

The mistake is in your calculation of $P(A_s)$. You forgot that the ace of spades can be picked on either turn, so $$P(A_s)=\frac1{52}+\frac {51}{52}\cdot\frac1{51}=\frac1{26}.$$
• 116k
1 vote
Accepted

### Density Functions and Probability Calculation

I don't need help solving this integral, I just wanna know how to get here. $$\int_{0.5}^1\int_{1.5-y}^11\,\mathrm d x\,\mathrm d y$$ The support is $\{(x,y):0\leq x\leq 1~,~ 0\leq y\leq 1\}$ and the ...
• 130k
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### Conditional Probability in Selection of Nuts from Different Weight Distributions

I will just find the probability of picking a nut from Bavaria (given its weight is >10) since you can then do the others. You must use Bayes theorem  P(B|X>10) = \frac{P(X>10|B)P(B)}{P(X&...
• 728
Accepted

### Conditional expectations preserve convergence in measure

Easy counter-example: Suppose $X$ and $X_n$ are independent of $\mathcal G$. Then you are asking if $X_n \to X$ a.s. implies $EX_n \to EX$. A standard counter-example is $X_n=n1_{(0,\frac1 n)}, X=0$ ...
• 38.2k
1 vote
Accepted

### Once $X\sim \text{gamma}(\alpha = 12, \beta = 2)$ is observed, $Y$ is randomly chosen from $(0,x)$. Evaluate $E(Y)$

From 'randomly chosen from $(0,x)$' we infer that, given any fixed value $x$ that $X$ can take, $Y$ follows a uniform distribution in $(0,x)$. This means that $E[Y\,|\,X=x]=\dfrac{x}{2}$, and so \$E[Y\,...
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