GATE Solved Paper 2017-19 - GATE 2018

47. Two people, P and Q, decide to independently roll two identical dice, each with 6 faces. Numbered 1 to The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by P and Q. Assume that all 6 numbers on each dice are equiprobable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins in the third trial is ___________.

Note – Numerical Type question

  • Option : C
  • Explanation :
    Given there are two identical dices, each having 6 faces.
    Favorable events for tie = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
    Since two dice are thrown, total sample space will be 6 x 6 = 36

    Therefore, P(tie) = 6/36 = 1/6
    and Probability of not tie = (1 – 1/6)
    To one of them win on third trial, previous two trials should be tie.

    = 1/6 * 1/6 * (1 – 1/6)
    = 1/36 * 5/6
    = 5/216 = 0.023
    So, option (C) is correct.
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48. The value of

correct to three decimal places (assuming that π = 3.14) is __________.

Note – Numerical Type question

  • Option : B
  • Explanation :
    Gate2018
    put x2 = t
    2xdx = dt or x dx = 1/2 dt
    if x = 0 then t = 0 and if x = π/4 then t = (π/4)2
    Gate2018
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49. The chromatic number of the following graph is __________.



Note – Numerical Type question

  • Option : A
  • Explanation :
    Sub graph is K3 chromatic number is at least 3. We can try for a chromatic number of 3 by using 3 colors, as follows :
    Gate2018 cs
    All vertices with only 3 colors, the chromatic number of this graph is 3.
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50. A 32-bit wide main memory unit with a capacity of 1 GB is built using 256M x 4-bit DRAM chips. The number of rows of memory cells in the DRAM chip is 214. The time taken to perform one refresh operation is 50 nanoseconds. The refresh period is 2 milliseconds. The percentage (rounded to the closest integer) of the time available for performing the memory read/write operations in the main memory unit is ____________.

  • Option : C
  • Explanation :
    Total number of rows is 214 and time taken to perform one refresh operation is 50 nanoseconds. So, total time taken to perform refresh operation = 214*50 nanoseconds = 819200 nanoseconds = 0.819200 milliseconds.
    Refresh period is 2 milliseconds. So, time spent in refresh period in percentage = (0.819200 milliseconds) / (2 milliseconds) = 0.4096 = 40.96%
    Hence, time spent in read/write operation = 100% – 40.96% = 59.04% = 59 (in percentage and rounded to the closet integer).
    So, answer is 59.
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