retagged by
10,831 views
35 votes
35 votes
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 ____
retagged by

8 Answers

1 votes
1 votes
P(Both dice have same number)=P(tie)= 1/6

P(Both dice have different number)=P(One of them wins)=5/6

The question was find the probability whether one of them wins in the third trial:  p(tie)*p(tie)*p(one of them wins)= 1/6*1/6*5*6= 0.023
1 votes
1 votes

Answer should be....0.0231
Prob(One of them wins on 3rd trial) Indirectly means first two trials were tie = prob ( Failure in 1st trial x failure in 2nd trial x success in 3rd trial )
Tie Or failure in our case happens= { (1,1), (2,2), (3,3), (4,4), (5,5), (5,5) } = which gives 1/6.
Therefore = 1/6 * 1/6*(1-1/6) = 0.0231.

Answer:

Related questions

32 votes
32 votes
6 answers
2
41 votes
41 votes
9 answers
4