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UGC NET CSE | Junet 2015 | Part 3 | Question: 64

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Given the symbols A, B, C, D, E, F, G and H with the probabilities$\frac{1}{30}, \frac{1}{30}, \frac{1}{30}, \frac{2}{30}, \frac{3}{30}, \frac{5}{30}, \frac{5}{30}$ and $\frac{12}{30}$respectively. The average Huffman code size in bits per symbol is

  1. $\frac{67}{30}$
  2. $\frac{70}{34}$
  3. $\frac{76}{30}$
  4. $\frac{78}{30}$
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ans is C  76/30

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A – 10100 – 5 bits
B – 10101 – 5 bits
C – 1100 – 4 bits
D – 1101 – 4 bits
E – 1011 – 4 bits
F – 100 – 3 bits
G – 111 – 3 bits
H – 0 – 1 bit
average Huffman code size = 5 * (1 / 30 ) + 5 * (1 / 30 ) + 4 * (2 / 30 ) + 4 * (3 / 30 ) + 3 * (5 / 30 ) + 3 * (5 / 30 ) + 1 * (12 / 30 ) = 76 / 30.
So, option (D) is correct.

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