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A real number y is chosen at random such that 0 <100. What is the probability that y-[y]=1/3?

 

Please help! I do not really know the exact way to calculate this, so please explain your reasoning! Thanks so much!

 Apr 24, 2023
 #1
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Is it supposed to be \(y - \lfloor y \rfloor \) or \(y + \lfloor y \rfloor \)?

 Apr 24, 2023
 #2
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It's \(y + \lfloor y \rfloor\)

Guest Apr 24, 2023
 #3
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Hello,

I would like to suggest you here

Let's first define what y-[y] means. This is the fractional part of y, which is the difference between y and the greatest integer less than or equal to y. For example, if y = 3.5, then [y] = 3 and y-[y] = 0.5. If y = 4.2, then [y] = 4 and y-[y] = 0.2.

Now, we want to find the probability that y-[y]=1/3, where 0 <100. Since y is chosen at random, we can assume that y is uniformly distributed on the interval (0,100). We can then break up this interval into subintervals of length 1/3, since we are interested in the case where y-[y]=1/3.

Let's consider the first subinterval, (0,1/3). For y to satisfy y-[y]=1/3 in this interval, we need y to be between 1/3 and 2/3. The probability of this happening is 1/3, since y is uniformly distributed on the interval (0,1).

Next, let's consider the subinterval (1/3,2/3). For y to satisfy y-[y]=1/3 in this interval, we need y to be between 2/3 and 1. The probability of this happening is also 1/3.

We can continue this process for the remaining subintervals, up to the interval (98 2/3, 99). For this interval, we need y to be between 98 2/3 and 99. The probability of this happening is again 1/3.

Therefore, the total probability that y-[y]=1/3 is the sum of the probabilities for each subinterval, which is:

1/3 + 1/3 + ... + 1/3 (98 times)

= 98/3

= 32.67%

So the probability that y-[y]=1/3 is approximately 32.67%. hope so it will help you  www.gmglobalconnect.com

 Apr 25, 2023
 #4
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The probability is 1/12.

 Apr 25, 2023

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