Does the Probability Question Involve Conditional Probability?

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The probability of rolling two dice, where the first die shows a perfect square and the second die shows a 4, is calculated as follows: the first die has a probability of 2/6 for being a perfect square (1 or 4), and the second die has a probability of 1/6 for being a 4. This results in a combined probability of (2/6) * (1/6) = 2/36, which simplifies to 1/18. The discussion highlights that if the question were rephrased to indicate that one die is a 4 and the other a perfect square, the probability would change to 3/36. The critical distinction lies in the wording and whether the events are treated as ordered or unordered.
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Two dice are rolled. What is the probability that the first die is a perfect square and the second die is a 4?


First die : 2/6

Second die : 1/6

Won't the probability be just (2/ 6) * (1 / 6) = 2 / 36 = 1 / 18

My answer says 3 / 6 but aren't these events not mutually exclusive? :|
 
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The wording is critical. If the question actually reads "that one is a four and the other a perfect square" then the answer is 3/36. This is because the order doesn't matter now, so the possible rolls are 1, 4; 4, 1; 4, 4. As soon as you discriminate the dice as first, second, it collapses to 2/36.
 
nesan said:
Two dice are rolled. What is the probability that the first die is a perfect square and the second die is a 4?


First die : 2/6

Second die : 1/6

Won't the probability be just (2/ 6) * (1 / 6) = 2 / 36 = 1 / 18

My answer says 3 / 6 but aren't these events not mutually exclusive? :|

Hi,

The sample space for your question is S={(i,j), i=1,..,6 j=1,...,6} and supposing that the dice are fair, all 36 outcomes have probability 1/36 of occurring.

E is the event: the first dice is a perfect square and the second is a 4.

E={(1,4),(4,4)}.

So P(E)=2/36=1/18 and not 1/2.

If we think about it, if we roll the dice let's say 100 times, we would not expect to get a perfect square for the first dice together with a 4 for the second dice around 50 times.

Does your question involve conditional probability?
 
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