Find one mode or two exactly tied modes

Let X count affected people among 12. Under the source constant-p independent-person model, $X\sim B(12,0.8)$. Enter Y1=binompdf(12,0.8,X), press 2nd, GRAPH, and scan the probabilities over the count support. The source retains these neighboring values, with a Teacher note to present this table:

The largest probability is at 10, so 10 people is the most likely count.

Let Y count affected people among 19, with $Y\sim B(19,0.8)$. Enter Y1=binompdf(19,0.8,X), press 2nd, GRAPH, and scan for the maximum. The Teacher completes the table and both answer blanks:

Both 15 and 16 are most likely.

The source self-reading note reports table readings 0.2181994019466 at 15 and 0.2181994019459 at 16, differing in the 12th/13th decimal places. The home screen prints 0.2181994019 for both, agreeing to 10 decimal places. Do not infer exact equality merely from nearly equal rounded output; retain the source formula check: $\dfrac{P(Y=15)}{P(Y=16)}=\dfrac{\binom{19}{15}(0.8)^{15}(0.2)^4}{\binom{19}{16}(0.8)^{16}(0.2)^3}=\dfrac{3876(0.2)}{969(0.8)}=1$. Thus the masses really are equal. The source’s practical advice to treat tiny table differences as numerical noise is justified here by this exact ratio, rather than as a universal equality test.