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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 ____
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I just want to know whether P and Q are rolling two identical dice, meaning total 4 dice or each are given one identical dice, meaning total 2 dice?
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why probability of tie 1/6 why not 5/6
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because there are 36 elements in the sample space and 6 of them are the same. $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and  (6, 6)$ out of $36$ elements. $6/36 = 1/6$.
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8 Answers

56 votes
56 votes
Best answer
One of them win in the third trial i.e. first two trial would be Tie and third should not be Tie.

Probability of Tie $=\frac{6}{36}=\frac{1}{6}$
Probability of NO Tie $=1-\frac{1}{6}=\frac{5}{6}$

Winning in the third Tie $= \text{(First Tie)} \ast \text{(Second Tie)} \ast \text{(No Tie)} = \frac{1}{6}\ast \frac{1}{6}\ast \frac{5}{6}= \frac{5}{216} =  0.023$
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Rishab bhai aapki to rank 1 aati fir
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@Digvijay Pandey @Mk Utkarsh why the cases of winning in frst and second trials is not added

i.e. why (5/6) + (1/6)*(5/6) + (1/6)*(1/6)*(5/6) is not correct here??

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because in question it is asking only about third trial. if asked that probability that one of them wins at most third trial.
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10 votes
10 votes
1st trail
Collision in first will have cases 1,1 2,2 3,3 4,4  5,5 6,6 So total 6 cases
Probability  = 6/36  = 1/6

2nd trail
Collision  will have cases 1,1 2,2 3,3 4,4  5,5 6,6 So total 6 cases
Probability  = 6/36  = 1/6

3rd trIal
Anyone of them, wins means no collision
So Probality = 1-1/6 = 5/6
So
As they all are independent cases so the answer is 1/6*1/6*5/6
0.023

1 comment

i think u mistyped trial for first two case
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8 votes
8 votes

Win in 3rd trial means tie in first 2.

In an event of rolling a die twice, there are 36 possible outcomes and out of which 6 of them make tie.

So, required probability=P(tie)*P(tie)*P(no tie) 

=(6*6*30)/(36*36*36)

=5/(6*6*6)=5/216

=0.0231 (ANS)

2 votes
2 votes
I Think it's 30/ (36)^2 = .0231
Answer:

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