+6 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!
+2666 Is it supposed to be \(y - \lfloor y \rfloor \) or \(y + \lfloor y \rfloor \)?
+2 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
