# Thread: Random choice of numbers from series, probability the numbers are equal?

1. ## Random choice of numbers from series, probability the numbers are equal?

A number X is chosen at random from the series 2,5,8.. and another number Y is chosen at random from the series 3,7,11... Each series has 100 terms. Find P[X=Y]

They way I solved this is as following.

I know that each series has 100 terms, so I can find the final term in each series.

Let $\displaystyle S_1=2,5,8..$and let $\displaystyle S_2=3,7,11..$

Therefore, the final term of $\displaystyle S_1$ should be 302 and the final term of $\displaystyle S_2$ should be 403.

Now I look look for the terms that both series share. I know that both series can't share any terms >302.

So I have

{11,23,35,47,59,71,83,95,107,119,131,143,155,167,1 79,191,203,215,227,239,251,263,275,287,299}

Or 25 terms shared between the two series.

Therefore, the probability of $\displaystyle P[X=Y]=\frac{25}{10,000}=\frac{1}{400}$

My question is whether there's an easier/faster way to solve this problem other than writing out and summing the number of terms? As you can see, it's not too bad if the series are rather short, but I'm interested in how I would approach this problem if each series were huge.

Thanks

2. ## Re: Probability question

The first time the numbers of both sequences are equal is for the 4th term of $\displaystyle S_1$
which is 11.

One sequence rises by 3 and the other by 4, so further equal numbers occur
for every multiple of 3(4)=12 added to 11.

That's every 4th term of $\displaystyle S_1$

The values of $\displaystyle S_1$ are smaller per position than $\displaystyle S_2$

The number of equal values is $\displaystyle \frac{100}{4}$