Probability
- Thread starter QKALH6679
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HI @QKALH6679,
To calculate the probability of observing an event in A, given that an event in C has been observed we use the following formula:
P(A|C) = P(A ∩ C) / P(C)
In this exercise, P(A ∩ C) is represented by the intersection between the two rectangles A and C (top-half of rectangle C) and its sum is 3% + 3% = 6%. At this point, we have calculated the numerator. We need to find the denominator P(C) which is represented by the whole rectangle C (3% + 3% + 3% + 3% = 12%). So, we are ready to calculate P(A|C) = 6% / 12% = 50%.
You need to follow the same steps to calculate P(B|C), the denominator is C rectangle with P(C) = 12% (same as the previous case) and the numerator is the intersection between B and C (left-half of rectangle C) 3% + 3% = 6%. The result is 6% / 12% = 50%.
I hope this answer helps you. In case you need more details please check out this thread where David explains it perfectly:
To calculate the probability of observing an event in A, given that an event in C has been observed we use the following formula:
P(A|C) = P(A ∩ C) / P(C)
In this exercise, P(A ∩ C) is represented by the intersection between the two rectangles A and C (top-half of rectangle C) and its sum is 3% + 3% = 6%. At this point, we have calculated the numerator. We need to find the denominator P(C) which is represented by the whole rectangle C (3% + 3% + 3% + 3% = 12%). So, we are ready to calculate P(A|C) = 6% / 12% = 50%.
You need to follow the same steps to calculate P(B|C), the denominator is C rectangle with P(C) = 12% (same as the previous case) and the numerator is the intersection between B and C (left-half of rectangle C) 3% + 3% = 6%. The result is 6% / 12% = 50%.
I hope this answer helps you. In case you need more details please check out this thread where David explains it perfectly:
thank you so much
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