Digital data received from a sensor can fill up 0 to 32 buffers. Let the…

Goal: find the probability that an even number of buffers are full.

Given p(j) = (1/562) (33 - j) for j = 0,1,...,32, the required probability is the sum of p(j) over even j.

Compute P(even) = (1/562) * sum_{even j=0}^{32} (33 - j).

• There are 17 even values from 0 to 32 (0,2,...,32). Write j = 2k for k = 0..16.

• Sum_{even j} (33 - j) = sum_{k=0}^{16} (33 - 2k) = 17*33 - 2*sum_{k=0}^{16} k.

• sum_{k=0}^{16} k = (16*17)/2 = 136, so the sum is 17*33 - 2*136 = 561 - 272 = 289.

Therefore P(even) = 289/562 ≈ 0.5146, which rounds to 0.515.

Note: The total of p(j) over j = 0..32 equals 561/562 (not 1) with the given formula. Using the probabilities as provided, the event probability computed above is 289/562 ≈ 0.515. If the distribution were intended to sum to 1, p(j) would need to be re-normalized (divide by 561 instead of 562).