Test started Apr 17, 2024 6:00 PMtoApr 24, 2024 6:00 PM
1.1
Question
A quiz consists of 10 multiple choice questions, each question has 4 options. You score +1 for each correct answer, and -0.5 for each incorrect answer.
Suppose you choose an answer uniformly at random for each question. What is your expected score?
Enter your answer to 2 d.p. (answers within ±0.01 will be marked correct)
Answer
对于每一道题,有0.25的概率答对,0.75的概率答错,因此每一道题的期望得分是:
E(X)=0.25×1+0.75×(−0.5)=−0.125
对于每一道题,它的期望都是一致的。因此,十道题的得分期望是:
E(X1+...+X10)=E(10×X1)=10×E(X1)=−1.25
1.2
Question
Suppose A, B, C, D are events with A⊥B, B⊥C and C⊥D.
True or false: It must be the case that D⊥A.
Answer
当事件 A, B, C, D 之间存在相互独立的关系时,如 A⊥B, B⊥C, C⊥D,这只意味着某些特定事件之间独立。独立性不具有传递性,即如果事件 A 独立于事件 B,并且事件 B 独立于事件 C,这并不意味着事件 A 必然独立于事件 C。
同理,尽管 B 独立于 C,且 C 独立于 D,我们不能推断出 A 必然独立于 D,除非有额外的信息表明这种独立性成立。这种独立关系不是通过中介事件自动传递的。因此,没有具体证据表明 A 和 D 独立,我们不能假设它们独立。
FALSE
1.3
Question
Suppose we have three urns with black and white balls distributed as follows:
- Urn A has 10 black balls and 20 white balls
- Urn B has 10 black balls and 10 white balls
- Urn C has 10 black balls and 1 white ball.
Suppose we choose an urn (uniformly at random) and draw a ball (uniformly at random) from that urn. What is the probability that the ball is white?
Answer
分别计算。
- 抽中A中的白球:
31×3020=92
- 抽中B中的白球:
31×2010=61
- 抽中C中的白球:
31×111=331
总概率即:
92+61+331=19883
1.4
Question
Let P be a probability distribution, and A and B events such that P(A) ,P(B), P(A|B) and P(B|A) are all non-zero.
True or false: P(A)=P(B) if and only if P(A∣B)=P(B∣A)
Answer
我们可以来进行一下推理。
P(A∣B)=P(B)P(A∩B)
P(B∣A)=P(A)P(B∩A)
由于 所有变量均不为0,因此充分性和必要性都得以证明。
TRUE
2.1
Question
True or false:
For any random variable X, the probability that X>E(X) is 1/2
Answer
X可以是非对称分布。
FALSE
2.2
Question
One way to simulate a die roll with a fair coin is as follows. Toss the coin three times:
- If the coin shows HHH: Output 1
- If the coin shows HHT: Output 2
- If the coin shows HTH: Output 3
- If the coin shows HTT: Output 4
- If the coin shows THH: Output 5
- If the coin shows THT: Output 6
- If the coin shows TTH or TTT: Repeat the process.
It was shown in lectures that each of the six outputs is equally likely, with probability 1/6.
A trial consists of three coin tosses. Note that if the coin shows TTH or TTT there will be more than one trial.
What is the expected number of trials before an output is produced?
Enter your answer to 3 d.p.s. An answer within ±0.001 will be correct.
Answer
我们先对这个期望进行计算。
E(Xn)=i=1∑∞i×86×(82)i−1=i=1∑∞i×43×(41)i−1=i=1∑∞4i3i=i=1∑∞4i−1×43(i−1)+3=i=1∑∞41×[4i−13(i−1)+4i−13]=41i=1∑∞4i−13(i−1)+41i=1∑∞4i−13=41E(Xn)+3×i=1∑∞4i1
这个几何级数 ∑i=1∞4i1 收敛于 31。因此,原式可表达为:
E(Xn)=41E(Xn)+3×31
计算得到:
E(Xn)=34
另外,GPT给出了一种更简单的证明思路:
因此,答案为:
1.333
2.3
Question
Suppose two normal, six-sided dice are rolled, a red die and a black die.
Let:
- A be the event that both dice show 4 or higher;
- B be the event that the red die shows 4 or lower; and
- C be the event that the black die shows 4.
True or false: A and (B∩C) are independent events.
Answer
判断是否为独立事件,最简单的方法就是判断它们同时发生的概率是否等于它们各自发生的概率的乘积,即:
P(A∩(B∩C))=P(A)×P(B∩C)
首先我们可以意识到,B和C确实是独立事件,它们是不同的骰子,因此互不影响。
P(A)=63×63=41
P(B)=64=32
P(C)=61
因此,我们可以得到:
P(B∩C)=32×61=91
同时,我们可以注意到,A、B、C同时发生时,需要两个骰子都显示4。
则:
P(A∩(B∩C))=61×61=361=P(A)×P(B∩C)
因此,答案为:
TRUE
2.4
Question
You are in a maze with 4 rooms. To navigate the maze you decide to wander randomly with the following strategy:
- In room A you will exit with probability 1/3, move to room B with probability 1/3, and move to room C with probability 1/3.
- In room B you will move to room D with probability 1/3, and room A with probability 2/3.
- In room C you will move to room D with probability 2/3, and room A with probability 1/3.
- In room D you will exit with probability 1/3, move to room B with probability 1/3, and move to room C with probability 1/3.
Starting from Room A, what is the probability that you will exit the maze from Room D?
Enter your answer to 3 d.p.s. An answer within ±0.001 will be correct.
Answer
我们设置:PX 为从房间 X 开始逃脱的概率。
我们直接将从A离开的可能性置为0(因为它没有去任何其他的房间),从D离开的可能性置为1(我们要找的)。
那么:
PA=31×0+31PB+31PC
PB=32PA+31PD
PC=31PA+32PD
PD=31×1+31×PB+31×PC
联立方程代入解得 PA=31=0.333