If a fair die with six sides is tossed 6 times, what is the probability that at least two of outcomes will
be the same?
probability (atleast 2) = 1 - 1/6 = 5/6
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Rahul@gurome
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For this problem, it is easier to calculate the complement probability of the event rather than the probability of the event itself.yellowho wrote:If a fair die with six sides is tossed 6 times, what is the probability that at least two of outcomes will be the same?
Probability that at least two outcomes are same = 1 - (Probability that all the outcomes are different)
Now, if we roll a six-sided dice 6 times, all the outcomes will be different if all possible outcomes occur, i.e. 1, 2, 3, 4, 5, and 6 in different rolls. Thus number of different sets of all different outcomes is nothing but the number of possible different arrangements of this 6 outcomes.
Thus, number of possible ways such that all the outcomes are different = 6! and total number of possible outcomes = 6^6
Probability that all the outcomes are different = 6!/(6^6)
Required probability = 1 - [6!/(6^6)]
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quite scientific, whereas the smallest probability here is 1/6. Two events will be the same if the probability is not 1/6 -> this is included with our condition of 'at least 2 the same'. Hence P(2 the same)=1-Probability(not all the same) or 5/6Rahul@gurome wrote:For this problem, it is easier to calculate the complement probability of the event rather than the probability of the event itself.yellowho wrote:If a fair die with six sides is tossed 6 times, what is the probability that at least two of outcomes will be the same?
Probability that at least two outcomes are same = 1 - (Probability that all the outcomes are different)
Now, if we roll a six-sided dice 6 times, all the outcomes will be different if all possible outcomes occur, i.e. 1, 2, 3, 4, 5, and 6 in different rolls. Thus number of different sets of all different outcomes is nothing but the number of possible different arrangements of this 6 outcomes.
Thus, number of possible ways such that all the outcomes are different = 6! and total number of possible outcomes = 6^6
Probability that all the outcomes are different = 6!/(6^6)
Required probability = 1 - [6!/(6^6)]
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yellowho
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Is there a different to do this? The OA is 319/324. It's exactly the same answer but that expression suggests that there's a different methodology. 6!/6^6 is quite the calculation.
[quote="Rahul@gurome"][quote="yellowho"]If a fair die with six sides is tossed 6 times, what is the probability that at least two of outcomes will be the same?[/quote]
For this problem, it is easier to calculate the complement probability of the event rather than the probability of the event itself.
Probability that at least two outcomes are same = 1 - (Probability that all the outcomes are different)
Now, if we roll a six-sided dice 6 times, all the outcomes will be different if all possible outcomes occur, i.e. 1, 2, 3, 4, 5, and 6 in different rolls. Thus number of different sets of all different outcomes is nothing but the number of possible different arrangements of this 6 outcomes.
Thus, number of possible ways such that all the outcomes are different = 6! and total number of possible outcomes = 6^6
Probability that all the outcomes are different = 6!/(6^6)
Required probability = 1 - [6!/(6^6)][/quote]
[quote="Rahul@gurome"][quote="yellowho"]If a fair die with six sides is tossed 6 times, what is the probability that at least two of outcomes will be the same?[/quote]
For this problem, it is easier to calculate the complement probability of the event rather than the probability of the event itself.
Probability that at least two outcomes are same = 1 - (Probability that all the outcomes are different)
Now, if we roll a six-sided dice 6 times, all the outcomes will be different if all possible outcomes occur, i.e. 1, 2, 3, 4, 5, and 6 in different rolls. Thus number of different sets of all different outcomes is nothing but the number of possible different arrangements of this 6 outcomes.
Thus, number of possible ways such that all the outcomes are different = 6! and total number of possible outcomes = 6^6
Probability that all the outcomes are different = 6!/(6^6)
Required probability = 1 - [6!/(6^6)][/quote]
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Rahul@gurome
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I believe the method is same as I explained.yellowho wrote:Is there a different to do this? The OA is 319/324. It's exactly the same answer but that expression suggests that there's a different methodology. 6!/6^6 is quite the calculation.
The calculation is not at all complicated as it looks. Proceed as follows,
- 6!/6^6 = (1*2*3*4*5*6)/(6^6) = (4*5*6*6)/(6^6) = (4*5)/(6^4) = 5/(3*3*6*6) = 5/324
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