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When Pam rolls two dice, there are total 36 (=6x6) cases.
For Example, (1,1), (1,2), (1,3), ..., (1,6), (2,1), (2,2), (2, 3), ..., (6,4), (6,5), (6,6).
Now, there are two scenarios:
Scenario 1: When both the dice show the same numbers.
There are 6 such cases: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
The probability that Pam rolls doubles is \(\frac{6}{36}.\)
Robin can also have 36 possibilities. But, Robin can match Pam in only 1 way just by rolling the same set of dice. That means out of 36 outcomes, there is only one winning outcome for Robin.
Therefore, the probability that Pam rolls doubles and Robin matches her outcome is \(\frac{6}{36} * \frac{1}{36}\)
Scenario 2: When the dice show different numbers.
There are 30 such cases: (1,2), (1,3), ..., (2,1), (2,3), ....(6,5).
The probability that Pam rolls two distinct numbers is \(\frac{30}{36}\).
Robin can also have 36 possibilities. But, Robin can match Pam in 2 ways. For example, if Pam gets (2,6), Robin can get the same outcome in 2 different ways - (2,6) and (6,2). That means out of 36 outcomes, there are exactly 2 winning outcomes for Robin.
Therefore, the probability that Pam rolls distinct and Robin matches her outcome is \(\frac{30}{36} * \frac{2}{36}\)
Therefore, total required probability = (\(\frac{6}{36} * \frac{1}{36}\)) + (\(\frac{30}{36} * \frac{2}{36}\)) = (\(\frac{1}{216})+(\frac{10}{216}) = \frac{11}{216}\).
For Example, (1,1), (1,2), (1,3), ..., (1,6), (2,1), (2,2), (2, 3), ..., (6,4), (6,5), (6,6).
Now, there are two scenarios:
Scenario 1: When both the dice show the same numbers.
There are 6 such cases: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
The probability that Pam rolls doubles is \(\frac{6}{36}.\)
Robin can also have 36 possibilities. But, Robin can match Pam in only 1 way just by rolling the same set of dice. That means out of 36 outcomes, there is only one winning outcome for Robin.
Therefore, the probability that Pam rolls doubles and Robin matches her outcome is \(\frac{6}{36} * \frac{1}{36}\)
Scenario 2: When the dice show different numbers.
There are 30 such cases: (1,2), (1,3), ..., (2,1), (2,3), ....(6,5).
The probability that Pam rolls two distinct numbers is \(\frac{30}{36}\).
Robin can also have 36 possibilities. But, Robin can match Pam in 2 ways. For example, if Pam gets (2,6), Robin can get the same outcome in 2 different ways - (2,6) and (6,2). That means out of 36 outcomes, there are exactly 2 winning outcomes for Robin.
Therefore, the probability that Pam rolls distinct and Robin matches her outcome is \(\frac{30}{36} * \frac{2}{36}\)
Therefore, total required probability = (\(\frac{6}{36} * \frac{1}{36}\)) + (\(\frac{30}{36} * \frac{2}{36}\)) = (\(\frac{1}{216})+(\frac{10}{216}) = \frac{11}{216}\).
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