Digital data received from a sensor can fill up 0 to 32 buffers. Let the…
2018
Digital data received from a sensor can fill up 0 to 32 buffers. Let the sample space be \(S=\{ 0, 1, 2, \dots , 32\}\) where the sample \(𝑗\) denote that \(𝑗\) of the buffers are full and \(p(i) = \frac{1}{562}(33-i)\). Let A denote the event that the even number of buffers are full. Then \(𝑃(𝐴) \) is
- A.
0.515
- B.
0.785
- C.
0.758
- D.
0.485
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Correct answer: A
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).
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