Final answer:
The probability that a marble drawn from a jar is either red or even-numbered is 5/7, after considering the probabilities individually and correcting for the overlap of a marble that is both red and even-numbered.
Step-by-step explanation:
The question asks us to find the probability that a marble drawn from a jar is either red or even-numbered. Given that the jar contains 3 red marbles numbered 1 to 3 and 4 blue marbles numbered 1 to 4, we will calculate the probability of drawing a red marble (P(Red)), and the probability of drawing an even-numbered marble (P(Even)).
To find P(Red), we see there are 3 red marbles out of 7 total marbles. Thus, P(Red) = 3/7.
To find P(Even), there are 3 even-numbered marbles out of 7 (the blue marbles numbered 2 and 4, and the red marble numbered 2). Thus, P(Even) = 3/7.
Since one marble is both red and even-numbered, we must subtract the probability of drawing that marble once to avoid double-counting. Hence, P(Red or Even) = P(Red) + P(Even) - P(Red and Even), which would be 3/7 + 3/7 - 1/7 = 5/7.
Therefore, the probability that a marble drawn is either red or even-numbered is 5/7.