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If a fair die is rolled twice and two numbers X1=result of the first roll and X2=result of the second roll are obtained. Given that X1+X2=7, what is the probability that X1 =4 or X2 =4

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Answer:

P = 2/3 = 0.67

Step-by-step explanation:

Assuming that X1 + X2 = 7

The total number of outcomes such that X1 + X2 = 7 are:

X1 = 3, X2 = 4

X1 = 4, X2 = 3

X1 = 1, X2 = 6

X1 = 6, X2 = 1

X1 = 2, X2 = 5

X1 = 5, X2 = 2

So there are 6 total outcomes such that X1 + X2 = 7

Now, the probability that X1 = 4 is equal to the quotient between the number of outcomes where X1 = 4 (only one outcome) and the total number of outcomes (6)

Then:

p = 1/6

And the probability that X2 = 4 is calculated in the same way, then the probability is:

q = 1/6

Now, the probability that X1 = 4 or X2 = 4 is the sum of both probabilities we've found above, then:

P = p + q = 1/6 + 1/6 = 2/6 = 2/3 = 0.67

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