If P(a<X<b) = 0.48 and P(0<X<a) = 3 x P(b<X<12), calculate P(b<X<12) and calculate P(0<X<b).
To calculate P(b<X<12), we can use the given information and the properties of probability.
First, we know that P(a<X<b) = 0.48. This means the probability of X being between a and b is 0.48.
We are also given that P(0<X<a) = 3 x P(b<X<12). Let's call P(b<X<12) as "p".
Using this information, we can set up the equation:
P(0 < X < a) = 3p
Since 0 < X < a implies 0 < X < 12 (as mentioned in the question), we can replace 0 < X < a with 0 < X < 12 in the equation:
P(0 < X < 12) = 3p
Now, we have another equation related to P(0 < X < 12).
Subtracting P(a < X < b) from P(0 < X < 12), we get:
P(0 < X < 12) - P(a < X < b) = 3p - 0.48
Since P(0 < X < 12) - P(a < X < b) is the same as P(0 < X < a) (as they refer to the same range of X), we can substitute it with the given value:
3p - 0.48 = 3 x P(b < X < 12)
Simplifying the equation, we get:
3p - 0.48 = 3p
This implies that 0.48 = 0, which is not possible.
Hence, there may be an error or inconsistency in the given information. Please double-check the values and equations provided.