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Sorry, for the two cases i=0 and 50, your expression is true. But I am afraid that your answer is still incorrect. Just simply check i=1. You have only one jump of 2 steps, and 98 jumps of 1 step. How many different ways? Should be 99, right? Because you can only start jumping 2 steps a time at...

1. For the square question, I reconsider a solution: E(n)=4. I think this should be the corret answer. One can only go back to the starting point by an even number of steps, say n=2k, for k=1,2,....infinity. For n=2k steps, all the odd steps do not matter, because they will lead to either B...

If I understand your idea correctly, the result should be the sum of (100-i)!/((i !)*(100-2i)!) for i=0,1,2...50, which is the number of jumps of 2 steps each. Supporse you have i jumps of 2 steps each, then you must have (100-2i) jumps of 1 step each. So to calculate the number of possible...

1. The same. Because how much milk is moved to the water bottle is always equal to the amount of water moved to the milk bottle, according to the conservation of the volume. 2. I can only rewrite is as a double integral, but cannot integrate it analytically. Any hints?? 3. This problem can...

P(xy<0.5)=0.8466. We can solve this easily using geometrical method. x,y are i.i.d uniform distributed in (0,1), so the sample space can be represented as a unit square. The curve xy=0.5 is easily determined which passes (1, 0.5) and (0.5,1). So P(xy<0.5) is the area of the region of the unit...

P(11)>P(12) On first sight, I guess P(11)=P(10) and has the maximum probability, so P(11)>P(12). I make this guess based on my knowledge about the distribution of the sum of two fair dice, which is easy to calculate. It has a symmetrical distribution with respect to P(7), which is the highest...

Some answers..... Here are my answers to some of the following questions: 1. The sum of all the digits must a multiple of 9. 4. Let En denotes the event that after n steps one returns back to the starting point for the first time. And I find P(En)=1/2 for all the n. Proceeding in thi...

To he yan's question: Following the Bayesian Theorem: P=(1/6*3/4)/(1/6*3/4+5/6*1/4)=3/8

To Andy`s original question: 1. Ans=sqrt(2). let x^x.....=I=2. From the recursive feature we can see x^x^x.....=x^I. so x^I=x^2=2 (because I=2). then x=sqrt(2). The same technique is used to solve 6. let left hand side=I. Then sqrt(2+I)=I. so 2+I=I^2, two solutions, but you can only...