probability wheels

There is $\frac{1}{4}$ chance the right wheel is 2. In which case there is a $\frac{1}{3}$ chance the sum will be 7 or greater.

There is $\frac{1}{4}$ chance the right wheel is 3. In which case there is a $\frac{2}{3}$ chance the sum will be 7 or greater.

There is $\frac{1}{4}$ chance the right wheel is 4. In which case there is a $\frac{5}{6}$ chance the sum will be 7 or greater.

There is $\frac{1}{4}$ chance the right wheel is 5. In which case there is a $100\%$ chance the sum will be 7 or greater.

Therefore, the total probability of a spin of 7 or greater is:

$\dfrac{\frac{1}{3} + \frac{2}{3} + \frac{5}{6} +1}{4} = \boxed{0.71}$

Let $p_n$ and $p'_n$ be the probability that the left and right disc, respectively, lands on $n \in \{2,3,4,5\}$ . Here, $p_2 = p_3=\frac{1}{6}, p_4 = p_5=\frac{1}{3}$ , and $p'_2= p'_3= p'_4= p'_5= \frac{1}{4}$ . The spin results from each disc are independent random events, so the probability that the left disc lands on $n$ and the right on $n'$ is $p_n p'_{n'}$ . The total probability that the sum of the results from both discs after each is spun is 7 or more is the sum of the individual probabilities of all the ways of achieving at least 7 total from both discs. Therefore: $\begin{aligned} P\{n+n' \ge 7\} &= p_2 p'_5 + p_3 (p'_4 + p'_5) + p_4 (p'_3 + p'_4 + p'_5) + p_5 (p'_2 + p'_3 + p'_4 + p'_5) \\ &= \frac{1}{6} \cdot \frac{1}{4} + \frac{1}{6} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{3}{4} + \frac{1}{3} \cdot 1 = \frac{17}{24} \approx\boxed{0.71} .\\ \end{aligned}$