UGC NET CSE | July 2018 | Part 2 | Question: 84

Question

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

Comments

Dimpu1211 i'm adding the answer here

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Answer: (a)

Answer 1

Given,

Probability of ith buffer getting full = $\large p(i) = \frac{1}{562}(33-i)$

Probability of all even number of buffers are full is $P(A)$

$P(A) =$$\large \sum_{i = 0,2,4,6,8,..,32} p(i) = \sum_{i = 0}^{16} p(2i)$

$P(A) =$ $\large \frac{1}{562} \left ( 33 + 31 + 29 + 27 + ....+ 1 \right )$

$P(A) =$$\frac{1}{562}\times 289 = 0.51423$

Answer 2

Sample Space j denotes that j of the buffers are full and probability of j = $\frac{1}{561} .$ * (33-j)       ( i hope it's a typing mistake )

P(A) = P(0) + P(2) +....P(32)

        = $\frac{1}{561} .$ * (33-0)  +  $\frac{1}{561} .$ * (33-0)  + ..... +$\frac{1}{561} .$ * (33-32)

        =  $\frac{1}{561} .$ * { (33-0) + ( 33-2) +....+(33-32) }

        = $\frac{1}{561} .$ * { 33 *17 - (0+2+4+6+...+32) }

        = $\frac{1}{561} .$ * { 33 *17 - (17*16) }

        = $\frac{17}{561} .$ * { 33-16 }

        = $\frac{17}{561} .$ * {17}

        = $\frac{289}{561} $ =  0.515

 

option A is correct