Digital data received from a sensor can fill up 0 to 32 buffers.
Digital data received from a sensor can fill up 0 to 32 buffers. Let the sample space be
S = {0, 1, 2, .........., 32} where the sample j denote that j of the buffers are full and \(p\left( i \right) = \frac{1}{{561}}\left( {33 - i} \right)\)
. 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
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Buffers are from 0 to 32
Sample space S = {0, 1, 2…………32}
Probability of ith buffer getting full is denotes by p(i) which is given as : \(p\left( i \right) = \frac{1}{{561}}\left( {33 - i} \right)\)
Probability of even number of buffers are full is denote by p(A).
We have to consider only the even numbers from the sample space i.e. 0, 2, 4, …………..32
For this, find the probability for p(0),p(2), p(4), p(6),……….p(32)
\(p\left( 0 \right) = \frac{1}{{561}}\left( {33 - 0} \right) = \;\frac{1}{{561}}\left( {33} \right)\)
\(p\left( 2 \right) = \frac{1}{{561}}\left( {33 - 2} \right) = \;\frac{1}{{561}}\left( {31} \right)\)
\(p\left( 4 \right) = \frac{1}{{561}}\left( {33 - 4} \right) = \;\frac{1}{{561}}\left( {29} \right)\)
…………….
\(p\left( {32} \right) = \frac{1}{{561}}\left( {33 - 32} \right) = \;\frac{1}{{561}}\left( 1 \right)\)
So, every time bracketed value is decreasing by 2 while outside value remains same.
So, p(A) = p(0 ) + p(2) + p(4) +…………….+ p(32)
P(A) = (1/561) [33 + 31 + 29 + 27 + 25 + 23 + 21 + ………….1]
p(A) = (1/561) × 289
p(A) = 0.515