A quick fix to Month Collision

The probability that on any bench of 5 students no two students have the same birth month is $\dfrac{12}{12} \times \dfrac{11}{12} \times \dfrac{10}{12} \times \dfrac{9}{12} \times \dfrac{8}{12} = \dfrac{55}{144}$ .

The probability that a bench has at least two students sharing a birth month is then $1 - \dfrac{55}{144} = \dfrac{89}{144}$ .

The probability that 4 or more of the 8 benches have students sharing birth months is thus

$\displaystyle \sum_{k = 4}^{8} \dbinom{8}{k} \times \left(\dfrac{89}{144}\right)^{k} \times \left(\dfrac{55}{144}\right)^{8 - k} \approx 0.853 \gt 0.5$ ,

implying that this scenario is indeed quite likely, i.e., the answer is $\boxed{\text{Yes}}$ .