Character | Probability |

P | 0.22 |

Q | 0.34 |

R | 0.17 |

S | 0.19 |

T | 0.08 |

Total | 1.00 |

If a message of 100 characters over X is encoded using Huffman coding, then the expected length of the encoded message in bits is_____.

Note – Numerical Type question

- Option : A
- Explanation :

By using haffman tree :

So, number of bit required for each alphabet:

T = 3 bit, R = 3 bit, Q = 2 bit, S = 2 bit, P = 2 bit

Then, average length per character is = (number of bits * frequency of occurance of each alphabets)

= 3 * 0.08 + 3 * 0.17 + 2 * 0.34 + 2 * 0.19 + 2 * 0.22 = 2.25 bits

And, average length for 100 character = 2.25 * 100 = 225 bits.

Hence, 225 bits is correct answer

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52. The next state table of a 2-bit saturating up-counter is given below.

Q_{1} | Q_{0} | Q_{1}^{+} | Q_{0}^{+} |

0 | 0 | 0 | 1 |

0 | 1 | 1 | 0 |

1 | 0 | 1 | 1 |

1 | 1 | 1 | 1 |

- Option : B
- Explanation :
Q _{1}Q _{0}Q _{1}^{+}Q _{0}^{+}T _{1}T _{0}0 0 0 1 0 1 0 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 0 0

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54. For any discrete random variable X, with probability mass function

- Option : B
- Explanation :

Derivative of g_{x}(z) evaluated at z=1 gives expectation E(X) of X.

Therefore, take derivative of gY(z) with respect to z, and plug in z=1

Derivative is N.β.(1 - β + βz)(N-1), plug in z=1, gives Nβ.

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55. If the ordinary generating function of a sequence

- Option : A
- Explanation :

= (1+Z) (1+3Z + 6Z^{2}+ 10Z^{3}+……….∞)

Using binomial theorem

= 1 + 4Z + 9Z^{2}+ 16Z^{3}+ …… ∞………(2)

From (1) and (2), a_{0}= 1 and a_{3}= 16

∴ a_{3}– a_{0}= 15

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