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Probability and Statistics

GATE CSE 2019 · Q22

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Two numbers are chosen independently and uniformly at random from the set {1,2,…,13}\{1, 2, \dots, 13\}. The probability (rounded off to 3 decimal places) that their 4-bit (unsigned) binary representations have the same most significant bit is _______.

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Solution

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Official answer: 0.503 (±0.001) (±0.001)

Probability and Statistics

The set of numbers is S={1,2,3,4,5,6,7,8,9,10,11,12,13}S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\}, so ∣S∣=13|S| = 13.

In 4-bit unsigned binary representation:

  • Numbers with most significant bit (MSB) equal to 00 are 11 to 77 (since 110=000121_{10} = 0001_2 to 710=011127_{10} = 0111_2). There are 77 such numbers.

  • Numbers with MSB equal to 11 are 88 to 1313 (since 810=100028_{10} = 1000_2 to 1310=1101213_{10} = 1101_2). There are 66 such numbers.

Two numbers are chosen independently and uniformly with replacement from the set.

  • Probability that both numbers have MSB = 00:

P(both 0)=(713)×(713)=49169P(\text{both } 0) = \left(\frac{7}{13}\right) \times \left(\frac{7}{13}\right) = \frac{49}{169}
  • Probability that both numbers have MSB = 11:

P(both 1)=(613)×(613)=36169P(\text{both } 1) = \left(\frac{6}{13}\right) \times \left(\frac{6}{13}\right) = \frac{36}{169}
  • Total probability of having the same MSB:

P=49169+36169=85169≈0.50296P = \frac{49}{169} + \frac{36}{169} = \frac{85}{169} \approx 0.50296

Rounding off to 3 decimal places gives 0.503 (acceptable range: 0.502 to 0.504).

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This Probability and Statistics previous year question is Q22 of the GATE CSE 2019 paper. Find more GATE CSE PYQs on Probability and Statistics in Engineering Mathematics, or browse all GATE CSE PYQs.

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Two numbers are chosen independently and uniformly at random from the set {1, 2, , 13}. The probability (rounded off to 3 decimal places) that their 4-bit (unsigned) binary representations have the same most significant bit is …

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