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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 _______ .
Note - This was Numerical Type question.
Note - This was Numerical Type question.
- a)0.6944
- b)0.1157
- c)0.023
- d)0.463
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Two people, P and Q, decide to independently roll two identical dice, ...
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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Two people, P and Q, decide to independently roll two identical dice, ...
Problem:
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 _______.
Solution:
To find the probability that one of the players wins on the third trial, we need to consider all possible outcomes and calculate the probability of each outcome.
Let's analyze the possible outcomes for the first two trials:
First Trial:
There are 36 possible outcomes when rolling two dice, as each die has 6 possible outcomes. Out of these, there are 6 outcomes where both players tie (1-1, 2-2, 3-3, 4-4, 5-5, 6-6), and 30 outcomes where one player wins and the other loses.
Second Trial:
In the second trial, there are two possible scenarios:
1. If there was a tie in the first trial, both players roll again. There are 6 possible outcomes for this scenario.
2. If there was a winner in the first trial, the loser of the first trial rolls again. There are 6 possible outcomes for this scenario.
Now, let's analyze the possible outcomes for the third trial:
Third Trial:
In the third trial, there are three possible scenarios:
1. If there was a tie in both the first and second trials, both players roll again. There are 6 possible outcomes for this scenario.
2. If there was a tie in the first trial but a winner in the second trial, the loser of the second trial rolls again. There are 6 possible outcomes for this scenario.
3. If there was a winner in the first trial, the game ends and there is no need for a third trial.
Calculating the Probability:
To calculate the probability of each scenario, we need to multiply the probabilities of each trial.
The probability of a tie in a trial is 6/36 = 1/6, as there are 6 tie outcomes out of 36 possible outcomes.
The probability of a win in a trial is 30/36 = 5/6, as there are 30 winning outcomes out of 36 possible outcomes.
Let's calculate the probability for each scenario:
Scenario 1: Tie in both the first and second trials
Probability = (1/6) * (1/6) = 1/36
Scenario 2: Tie in the first trial but a winner in the second trial
Probability = (1/6) * (5/6) = 5/36
Scenario 3: Winner in the first trial
Probability = (5/6)
To find the probability of one of them winning on the third trial, we need to calculate the sum of probabilities for scenarios 1 and 2:
Total Probability = Probability of Scenario 1 + Probability of Scenario 2
= 1/36 + 5/36
= 6/36
= 1
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 _______.
Solution:
To find the probability that one of the players wins on the third trial, we need to consider all possible outcomes and calculate the probability of each outcome.
Let's analyze the possible outcomes for the first two trials:
First Trial:
There are 36 possible outcomes when rolling two dice, as each die has 6 possible outcomes. Out of these, there are 6 outcomes where both players tie (1-1, 2-2, 3-3, 4-4, 5-5, 6-6), and 30 outcomes where one player wins and the other loses.
Second Trial:
In the second trial, there are two possible scenarios:
1. If there was a tie in the first trial, both players roll again. There are 6 possible outcomes for this scenario.
2. If there was a winner in the first trial, the loser of the first trial rolls again. There are 6 possible outcomes for this scenario.
Now, let's analyze the possible outcomes for the third trial:
Third Trial:
In the third trial, there are three possible scenarios:
1. If there was a tie in both the first and second trials, both players roll again. There are 6 possible outcomes for this scenario.
2. If there was a tie in the first trial but a winner in the second trial, the loser of the second trial rolls again. There are 6 possible outcomes for this scenario.
3. If there was a winner in the first trial, the game ends and there is no need for a third trial.
Calculating the Probability:
To calculate the probability of each scenario, we need to multiply the probabilities of each trial.
The probability of a tie in a trial is 6/36 = 1/6, as there are 6 tie outcomes out of 36 possible outcomes.
The probability of a win in a trial is 30/36 = 5/6, as there are 30 winning outcomes out of 36 possible outcomes.
Let's calculate the probability for each scenario:
Scenario 1: Tie in both the first and second trials
Probability = (1/6) * (1/6) = 1/36
Scenario 2: Tie in the first trial but a winner in the second trial
Probability = (1/6) * (5/6) = 5/36
Scenario 3: Winner in the first trial
Probability = (5/6)
To find the probability of one of them winning on the third trial, we need to calculate the sum of probabilities for scenarios 1 and 2:
Total Probability = Probability of Scenario 1 + Probability of Scenario 2
= 1/36 + 5/36
= 6/36
= 1
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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 _______ .Note -This was Numerical Type question.a)0.6944b)0.1157c)0.023d)0.463Correct answer is option 'C'. Can you explain this answer?
Question Description
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 _______ .Note -This was Numerical Type question.a)0.6944b)0.1157c)0.023d)0.463Correct answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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 _______ .Note -This was Numerical Type question.a)0.6944b)0.1157c)0.023d)0.463Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 _______ .Note -This was Numerical Type question.a)0.6944b)0.1157c)0.023d)0.463Correct answer is option 'C'. Can you explain this answer?.
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 _______ .Note -This was Numerical Type question.a)0.6944b)0.1157c)0.023d)0.463Correct answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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 _______ .Note -This was Numerical Type question.a)0.6944b)0.1157c)0.023d)0.463Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 _______ .Note -This was Numerical Type question.a)0.6944b)0.1157c)0.023d)0.463Correct answer is option 'C'. Can you explain this answer?.
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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 _______ .Note -This was Numerical Type question.a)0.6944b)0.1157c)0.023d)0.463Correct answer is option 'C'. Can you explain this answer?, a detailed solution for 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 _______ .Note -This was Numerical Type question.a)0.6944b)0.1157c)0.023d)0.463Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of 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 _______ .Note -This was Numerical Type question.a)0.6944b)0.1157c)0.023d)0.463Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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