GATE Computer Science (CS) 2018 Shift 1 Solved Paper

Question : 22 of 65

Marks: +1, -0

Two people. P and Q, decide to independently roll two identical dice, each with 6 faces. Numbered 1 to 6 . 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 equi-probable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is ───────.

Your Answer:

Validate

Solution:

P (one of them wins is 3rd trial) = P ( Ist trial is Tie ) × P( IInd trial is Tie ) × P (one of them wins 3rd trial )

{P}

(tie in any trial

)= {P}( {P}=1

and

{Q}=1)+ {P}( {P}=2

and

{Q}=2)+. . .+ {P}( {P}=6

and

{Q}=6

)

=\;{1}/{36}+\;{1}/{36}+\;{1}/{36}+\;{1}/{36}+\;{1}/{36}+\;{1}/{36}

=\;{6}/{36}=\;{1}/{6}

{P}

(one of them wins)

=1- {P}(

Tie

)

=1-\;{1}/{6}=\;{5}/{6}

So required probability

=\;{1}/{6} × \;{1}/{6} × \;{5}/{6}=\;{5}/{216}=0.023

(rounded to 3 decimal places)

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