388

The probability of not being poisoned is exactly the same as the following problem: You choose one cup and drink from the other three. What is the probability of choosing the poisoned cup (and not being poisoned)? That probability is 1/4. Therefore, the probability of being poisoned is 3/4.


267

Here's a pragmatic answer from an engineer. You can always get a fair 50/50 outcome with any "coin" (or SD card, or what have you), without having to know whether it is biased, or how biased it is: Flip the coin twice. If you get $HH$ or $TT$, discard the trial and repeat. If you get $HT$, decide $H$. If you get $TH$, decide $T$. The only conditions ...


213

For any individual chick, there is a $0.5$ chance that the one on its right won't peck it, and a $0.5$ chance that the one on its left won't peck it. So overall, it has a $0.25$ chance of not being pecked. Since this is true for every chick, we can add up $0.25(100)=25$ to get the number of unpecked chicks. "But wait," you might say, "the probabilities ...


188

You should never bet on that kind of sequence. Now, every poster will agree that the odds of any sequence from 000000000 through 999999999 has an equal probability. And if the prize is the same for all winners, it's fine. But, for shared prizes, you will find that you just beat 10 million to 1 odds only to split the pot with dozens of people. To be clear, ...


179

The question is very easy to answer without computing probabilities. Every combination with six different numbers contains exactly one six. There are then additional combinations which contain exactly one six - e.g. $111116$. So the probability of exactly one six is greater.


173

The most reasonable way to match the answer in the book would be to define the likelihood to be the ratio of success over failure (aka odds): $$ q=\frac{p}{1-p} $$ then the probability as a function of the odds is $$ p=\frac{q}{1+q} $$ In your case the odds are $4:1$ so $4$ times as likely would be $16:1$ odds which has a probability of $$ \frac{16}{17}=94....


157

The probability is $\frac{1}{2}$ because the last flip determines it.


154

There is no such thing as a uniform distribution on the plane. Without specifying how the points are chosen, the question is not properly stated. However, if the points are chosen independently from some continuous distribution (absolutely continuous with respect to Lebesgue measure), the probability of the third point lying exactly on the line through the ...


152

You are right to think of the probabilities as areas, but the set of points closer to the center is not a triangle. It's actually a weird shape with three curved edges, and the curves are parabolas. The set of points equidistant from a line $D$ and a fixed point $F$ is a parabola. The point $F$ is called the focus of the parabola, and the line $D$ is ...


151

Why do most probability graphs show a bell curve? As you suspect, there is a natural tendency for distributions to be bell-shaped. There are some distributions that are not bell-shaped at all. For example, the outcome of a roll of one fair die is a discrete uniform distribution: By IkamusumeFan - Own work This drawing was created with LibreOffice Draw, ...


144

Well, let there be two players $A$ and $B$. Let them flip $7$ coins each. Whoever gets more tails wins, ties are discounted. It's obvious that both players have an equal probability of winning $p=0.5$. Now let's extend this. As both players have equal probability of winning the first seven tosses, I think we can discard them and view the 8th toss as a ...


142

I decided to make an answer out of my comment, just for the heck of it. $n$ doors, $k$ revealed Suppose we have $n$ doors, with a car behind $1$ of them. The probability of choosing the door with the car behind it on your first pick, is $\frac{1}{n}$. Monty then opens $k$ doors, where $0\leq k\leq n-2$ (he has to leave your original door and at least one ...


142

You can use symmetry here - Starting at $5$, it is equally likely to get to $0$ first or to $10$ first. Now, if you get to $10$ first, then it is equally likely to get to $0$ first or to $20$ first. What does that mean for the probability of getting to $0$ before getting to $20$?


138

It is a fair game, so your expected value at the end has to be $5$ like you started. You must have $\frac 34$ chance to go broke and $\frac 14$ chance to end with $20$.


136

Your intuition is wrong. Compare the two statements A. The event "the lucky number has all its digits repeated" is much less probable than the event "the lucky number has a few repeated digits" B. The number 1111111 (which has all its repeated digits) is much less probable than the number 8174249 (which has a few repeated digits). A is true, B is false. ...


130

The product of two Gaussian random variables is distributed, in general, as a linear combination of two Chi-square random variables: $$ XY \,=\, \frac{1}{4} (X+Y)^2 - \frac{1}{4}(X-Y)^2$$ Now, $X+Y$ and $X-Y$ are Gaussian random variables, so that $(X+Y)^2$ and $(X-Y)^2$ are Chi-square distributed with 1 degree of freedom. If $X$ and $Y$ are both zero-...


125

If $A$ is a $2\times 2$ matrix with determinant $d$ and trace $t$, then the characteristic polynomial of $A$ is $x^2-tx+d$. If this polynomial has distinct roots (over $\mathbb{C}$), then $A$ has distinct eigenvalues and hence is diagonalizable (over $\mathbb{C}$). In particular, if $d$ and $t$ are such that the characteristic polynomial has distinct roots,...


124

A Bernoulli random variable has two possible outcomes: $0$ or $1$. A binomial distribution is the sum of independent and identically distributed Bernoulli random variables. So, for example, say I have a coin, and, when tossed, the probability it lands heads is $p$. So the probability that it lands tails is $1-p$ (there are no other possible outcomes for ...


118

Yes, that's fine. Events are mutually exclusive if the occurrence of one event excludes the occurrence of the other(s). Mutually exclusive events cannot happen at the same time. For example: when tossing a coin, the result can either be heads or tails but cannot be both. $$\left.\begin{align}P(A\cap B) &= 0 \\ P(A\cup B) &= P(A)+P(B)\\ P(A\mid B)&...


117

The most famous counter-intuitive probability theory example is the Monty Hall Problem In a game show, there are three doors behind which there are a car and two goats. However, which door conceals which is unknown to you, the player. Your aim is to select the door behind which the car is. So, you go and stand in front of a door of your choice. At this ...


114

NicNic8 has provided a nice intuitive answer to the question. Here are three alternative methods. In the first, we solve the problem directly by considering which cups are selected if she is poisoned. In the second, we solve the problem indirectly by considering the order in which the cups are selected if she is not poisoned. In the third, we add the ...


109

Depends on how good you are The explanation is intuitive: If you are not very good (probability that you make a single shot - p < 0.6), then your overall probability is not very high, but it is better to bet that you'll make 2 out of 4, because you may do it just by chance and your clumsiness has less chance to prove in 4 than in 6 attempts. If you are ...


108

The complement of at least one boy is all three girls So, $P($ at least one boy$)=1-P(GGG)$ $=\displaystyle1-\left(\frac12\right)^3$ This is the de facto way of solving problems of Probability of at least one in case of Binomial Distribution like tossing a coin etc.


106

Let $e$ be the expected number of tosses. It is clear that $e$ is finite. Start tossing. If we get a tail immediately (probability $\frac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\frac{1}{4}$), then the expected number is $e+2$. Continue $\dots$. If we get $4$ heads then a tail, the expected number is $e+5$. ...


104

First of all, you must understand that there is no such thing as a perfectly fair coin, because there is nothing in the real world that conforms perfectly to some theoretical model. So a useful definition of "fair coin" is one, that for practical purposes behaves like fair. In other words, no human flipping it for even a very long time, would be able to ...


104

You are mistaken in thinking that what you perceive as "the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science" is really "the massive importance that is afforded to Bayes' theorem in undergraduate courses in probability and popular science." But it's probably not your fault: ...


103

The contrast between your interpretation and that in the other answers points out that there is an ambiguity in the wording of the question, in particular in the use of "one". You interpret it as "the first". The other answerers interpret it as "at least one". Someone might interpret it as "exactly one" in which case the answer would be 0. To elaborate on ...


101

Can we infer that there exist two numbers separated by a gap of $10000$, such that no number in between them is prime? We can infer this regardless of what you wrote. For every gap $n\in\mathbb{N}$ that you can think of, I can give you a sequence of $n-1$ consecutive numbers, none of which is prime. There you go: $n!+2,n!+3,\dots,n!+n$. So there is no ...


98

By not switching, you win a car if and only if you chose correctly initially. This happens with probability $\frac{1}{4}$. If you switch, you win a car if and only if you chose incorrectly initially, and then of the remaining two doors, you choose correctly. This happens with probability $\frac{3}{4}\times\frac{1}{2}=\frac{3}{8}$. So if you choose to switch, ...


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