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35

The answer depends on the details of just how you obtained your information. If somebody who saw both dice gives you as sole information that at least one of them is$~6$, then that limits the possible outcomes to $11$ (out of originally $36$) possibilities, just one of which is double sixes. The probability you asked for would then be$~\frac1{11}$ as in the ...


6

The possibility of one die rolling a six is totally unrelated to the other die. Imagine rolling one die after the other. I think we all agree it's the same as rolling them at the same time. The first one can be 1,2,3,4,5,6 so the chance is 1/6. The second one can also be 1,2,3,4,5,6, this is totally unrelated to what we rolled with the first die, so the ...


4

If you have $N$ possible 'dates' and $k$ people, here's one way to get a good estimate for the probability of a collision, especially if $N$ is much larger than $k$: There are $\begin{pmatrix} k\\2 \end{pmatrix}=\frac{k^2-k}{2}$ pairs of people. The probability that any given pair of people has different birthdays is $\frac{N-1}{N}$. Thus the probability ...


4

Let $D$ denote the number of possible days. The probability of a collision for $x$ birthdays is $$1-\prod_{k=1}^{x-1}\left(1-\frac{k}D\right).$$ If $x^2\ll D$, this is roughly $$\frac{x^2}{2D}.$$ If $x=10^9$ and $D=2^{160}\approx10^{48}$ one gets approximately $$10^{-30}.$$


3

This problem needs a change in notation. $p$ and $q$ look like probabilities and hard to remember which is red and which is green. So, we have a box with $r$ red balls and $g$ green balls. We remove balls one at a time (at random) until there is one ball left. Show that the probability that the last ball in the box is red is $r/(r+g).$ Let the pair ...


3

Note the following things: If you extract $2$ white balls, the amount of black balls remains the same. If you extract $2$ black balls, the amount of black balls is reduced by $2$. (Duh!) If you extract one of each you add one black after it, so the amount of black balls remains the same. This means that the parity of the amount of black balls is ...


3

There are two ways to pose the condition of the question, which lead to different answers: (1) I happen to catch sight of the first-rolled die; it shows a $6$. (2) I know only that at least one of the dice is a $6$. In the first case, the answer is 1/6; in the second, it is 1/11. If we modify the condition of case 1, to specify the second-rolled die ...


3

There are a total of $36$ outcomes, consisting of ordered pairs $(a,b)$ where $a,b\in \{1,\cdots,6\}$. Of those outcomes, the following 11 are possible given your knowledge: $(1,6),\cdots,(6,6)$ and $(6,1),\cdots,(6,5)$ (we already counted $(6,6)$). Among those 11, there is 1 corresponding to both 6's. Thus your answer is $1/11$.


2

If the intuition is not yet clear, perhaps one can do a formal conditional probability calculation. Let $A$ be the event "at least one $6$" and $D$ the event "double $6$." We want $\Pr(D|A)$. By the definition of conditional probability this is $\frac{\Pr(A\cap D)}{\Pr(A)}$. The event $A\cap D$ is just the event $D$, and has probability $\frac{1}{36}$. ...


2

Number of ways of choosing $5$ students = $135 \choose 5$ Number of ways of choosing $5$ high-scoring students = $44 \choose 5$ Can you see where to go from here?


2

Edit: My original answer is shown below the line. Andre Nicolas has explained in the comments why it is incorrect. The correct answer to the first question is $\frac{15}{40}$ as can be seen if we imagine picking 15 balls out of 40 balls in an urn. We assume that each patient is equally likely to be chosen to be visited by the doctor for a given week. ...


2

Hint: Use $P(A\cup B) = P(A) + P(B) - P(A)P(B)$


2

In general, $$ P(A\,\,\text{or} \,\, B)=P(A)+P(B)-P(A\,\,\text{and} \,\, B) $$ for any events $A$ and $B$. So in order to calculate $P(A\,\,\text{or} \,\, B)$ we generally need all three quantities on the right. However, when $A$ and $B$ are independent, then, by definition, $$ P(A\,\,\text{and}\,\, B)=P(A)P(B) $$ and thus we only need $P(A)$ and $P(B)$ to ...


1

Yes. Yes, due to the collider at $C$. It seems like you're already familiar with $d$-separation. Your statement about Bayes's Theorem is correct. I do not understand what you mean in 3.1. Option A is incorrect, and Option B is correct. Option B comes from $$P(C \mid B) = P(C \cap A \mid B) + P(C \cap \overline{A} \mid B),$$ which is due to marginalization. ...


1

The condition means that if you continue extracting balls, the last one will be red. There are $\displaystyle p+q\choose\displaystyle p$ ways to order all the balls. If we fix the last one red, we can order the rest of them in $\displaystyle p+q-1 \choose \displaystyle p-1$ ways. The probability is then: $$\frac{p+q-1 \choose p -1}{p+q \choose p} = ...


1

For part b), recall that two events are independent iff $P(B \cap C) = P(B)P(C)$. In this case $P(B \cap C) = 4/15$ and $P(B)P(C) = 9/15*4/15$ which is not equal to $P(B \cap C)$. Therefore the events are NOT independent. So you are correct. Edit: We get the relationship $P(A \cap B) = P(A)P(B)$ to prove independence between two variables from $P(A|B) = ...


1

I formulate two problems here and give an answer to both of them. It is up to you to sort out wich of the problems you are dealing with. Personally I think it is problem 1. This because of its nice answer $\frac{p}{p+q}$. Draw all $n:=p+q$ balls and let $C_{i}\in\left\{ R,G\right\} $ denote the colour of the $i$th ball drawn. Problem1: To find is the ...


1

An urn contains $p$ red and $q$ green balls. Balls are drawn one by one till balls left in the urn are all red. Prove that the probability of this event is $p/(p+q)$ . Please note that you are told that balls of same colour are left in the urn and you need to find the probability that these remaining balls are red. No, just look to the ...


1

Just perform the computation using: $R(t) = p \{\omega | T(\omega) > t \} = \int_t^\infty f(x) dx$. It may help to draw $f$ and think about what $R(t)$ means in terms of the graph of $f$. It should be clear that no devices fail for $t<a$, and that all devices fail for $t>b$. This tells you what $R(t)$ is for these ranges. Remember that $f$ is a ...


1

We are presumably expected to assume that $X$ and $Z$ are independent. That really should have been explicitly specified. We want to calculate $E(XZ^2)-E(XZ)E(Z)$. By independence $E(XZ)E(Z)=E(X)E(Z)E(Z)=1/8$. The first term is $E(X)E(Z^2)$. But $E(Z^2)=1/2$ because $Z^2=Z$, Thus $E(XZ^2)=1/4$. Our covariance is therefore $1/4-1/8$.


1

Your conclusion is still wrong. Consider decreasing function $f(y)=e^{-y},\ w=(1,1),\ w'=(2.5,0)$ and probability density $X\sim e^{-X}\mathbf 1_{X>0}$. One sees $$E\big[f\big(\sum w_iX_i\big)\big]=\frac{1}{4}<\frac{1}{3.5}=E\big[f\big(\sum w'_iX_i\big)\big],$$ contrary to your claim.


1

Far less elegant, but you can view the string of $18$ cards as an ordered list. Let $$P(n,k)=\frac{n!}{(n-k)!}.$$ Then there are $P(36,18)$ possible orders in which the $18$ cards may be taken ($36$ choices for the first card, $35$ for the second, and so on...). In any of these possible outcomes, we are interested in having $4$ aces and $14$ non-aces. There ...


1

Recall that "median" means "50th percentile", meaning that you have a 50% chance of falling under the median. Also note that $\int_0^\infty f(t) dt$ = 1, so the function you wrote is a probability density function. This means that $P(t<T) = \int_0^T f(t) dt$


1

Hint: The median $a$ for a random variable $X$ with density is the unique number satisfying $$ P(X\leq a)=P(X\geq a)=\tfrac12. $$


1

Probability is another term for relative frequency. If I repeat an experiment $n$ times and observe that a chosen event occurs $k$ times then its probability $p(n)$ is $\frac{k}{n}$. Due to the law of large numbers $p(n)$ converges against a value $p$. Then $p$ is the value we tend to speak of when we refer to the probability of an experiment and a chosen ...


1

Simple answer: 1/6. The probability of one dice does not affect the other. That 1 was a 6 can be ignored; it happens to be true in this case, but is the same as saying "I saw that the wall of the room we were in was orange; given the wall was orange, what is the probability of the dice being a six?". So assuming the dice are regular 6 sided, unbiased die, ...


1

We want to choose exactly $1$ of the $2$ "good" objects and exactly $2$ of the $28$ "bad" objects. Hence, we obtain: $$ \frac{\binom{2}{1} \binom{28}{2}}{\binom{30}{3}} = \frac{27}{145} $$


1

Pick up the sticks an join the first picked up with the second, the third picked up with the fourth, et cetera. The probality that the second picked up is the original match of the first is $\frac{1}{2n-1}$. If this has occurred then we go on with $2n-2$ sticks. This leads to a probability of ...


1

The committee RAM competition knows from experience that the probability of successfully Contest is 0.95 for the student who has grade "very good" in BAC test , 0.5 one who has 'Good' in BAC test and 0.2 for others. He believes, moreover, that among the candidates to contest RAM 2014, 35% grade "very good" and 50% were marked "Good". •If we consider ...


1

The mean value of a positive random variable is always $ \ge 0$



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