# Tag Info

### Probability of drawing two balls from a bag

You can solve it both by using the conditional probability and independence. The probability of selecting the first ball as red is obviously $$P(r_1) = \frac{4}{9}.$$ Then for each of this selection ...

### Probability of drawing two balls from a bag

You are using the general multiplication rule, so you are actually computing $P(A)*P(B|A),$ as the two events are not independent, as you have suspected. That is why you reduced the denominator for ...
• 30.7k
Accepted

### $P[A > B, A > C]$, where $A$, $B$, and $C$ are i.i.d. random variables

This is the simplest example I could think of. If $A$, $B$, and $C$ are independent coin tosses with heads 1 and tails 0 then $$P(A>B, A>C)=1/8$$ since this only happens if A=1 and B and C are ...
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### $P[A > B, A > C]$, where $A$, $B$, and $C$ are i.i.d. random variables

Does the independence of A, B, and C imply the independence of the events A>B and A>C? No. The event of $A>B$ is evidence that $A$ is large, thus making it more likely that $A>C$ too. ...
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1 vote
Accepted

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### Determining CDF of a function with variable inside absolute value

Because if $y \geq 1$, $F_X(-y) = 0$, so it might be confusing to answer both intervals in the same manner. Indeed, you could take $F_X(y) - F_X(-y)$ in the whole interval [0,2], noting that in the ...
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### Probability of a white ball chosen from a bag if there are two bags.

I used a tree diagram. You draw all the options and their respective probabilities, then you can multiply them to get the probability for the event at the bottom of the tree. Here, getting White from ...
1 vote

### Probability of a white ball chosen from a bag if there are two bags.

It's easier not to use Bayes thereom to calculate $P(W|A)$. You are being confused by the notation of $|A$. But that just says if your universe of options is pulling marbles from a bag that is $A$ ...
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### Probability of a white ball chosen from a bag if there are two bags.

Firstly, for probability of drawing white given A, don't use P(W\A), the correct symbol is $P(W|A)$ And P(W|A) is simply the ratio of white balls to total balls in $A$, so there is nothing to prove ...
• 30.7k

### Probability that one of the tickets is number 7 and none number 18?

You need to be very careful here. The simplest way would be to compute it in the typical hypergeometric way, viz $$\dfrac{\binom11\binom{18}2}{\binom{20}3} =\; \frac{51}{380}$$ If you wish to go by ...
• 30.7k
1 vote
Accepted

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### Lower bound on tail when $\mathbb E[X] = \infty$
I may be misunderstanding your question, but if all you know is that $E[X]=\infty$, then the only lower bound on $P(X>\tau)$ is zero. Pick large $\tau>0$ and small $\varepsilon>0$ and ...