my class followed that Pr(W=k) must be greater than Pr(W=k+1) by using the logic that as in terms of binomial Pr(X=1) has a higher prob than Pr(X=2) etc
thus 5(k+1) > 50 - k
solve for k and get 15/2 which is 7.5 and must check whether k=8 or k=7 (since discrete) satisfies Pr(W=k) > Pr(W=k+1); would find k=8 satisfies whilst k=7 does not
Starting by solving 5(k+1)>50-k and testing for 7 and 8 is more valid than evaluating 50/6 = 8.333 and testing for 8 and 9. If you take X~Bi(5,0.598), for example, 5*0.598 gives you an expected value of 2.99. But, as it turns out, Pr(X=4)>Pr(X=2). So, while the 50/6 method would indicate you to test for 2 and 3 as 2<E(x)<3, you would ACTUALLY need to test for 3 and 4, if you had derived an equation for Pr(X=k+1)/Pr(X=k) with those values. For this specific question, Pr(W=9)>Pr(W=7), but you can't prove that merely by the fact that 50/6 is closer to 9.
Either way, you could technically start by just guessing that k is around 8. As long as you show that Pr(W=9)<Pr(W=8) AND Pr(W=7)<Pr(W=8), that's enough to show Pr(W=8) has the greatest value. So it doesn't really matter how you get there regardless.