Answer:
53/63
Step-by-step explanation:
Since you're spinning both spinners once, the maximum amount of times you could get an odd number is 2. That is if both spinners show an odd number.
One way to go about this problem is to sum all of the probabilities that have one odd outcome in them. That is if the first spinner shows odd and the second shows even, or if the first spinner shows even and the second shows odd, or if they both show odd.
However, this practice is very cumbersome and becomes annoying as soon as more spinners come into play. It will get you the result you're looking for, but it's not recommended and certainly is not efficient.
A different, more efficient approach to this problem is to find the probability that neither of the spinners show an odd number. Since all probabilities in a tree diagram add up to 1, then the probability of getting at least one odd result is the difference between 1 and the probability of not getting odd results at all.
[tex]P(\text{only evens}) = P(\text{first even}) \times P(\text{second even}) = \frac29 \times \frac57 = \frac{10}{63}\\\\\to P(\text{at least one odd}) = 1 - P(\text{only evens}) = 1 - \frac{10}{63} = \frac{63}{63} - \frac{10}{63} = \frac{63-10}{63} = \frac{53}{63}[/tex]
The probability of getting a minimum of one odd number in a game is 53/63.
