Seven dice are thrown 729 times. How many times do you expect at least four dice to show 3 or 5?

Probability of getting 3 or 5 in a single toss = $\cfrac{1}{6}+\cfrac{1}{6}=\cfrac{2}{6}=\cfrac{1}{3}$

$\therefore$Binomial Distribution with n=7, p=1/3 and q=2/3

N=729, r=4

$P(X=r)=^{n}C_{r}p^{r}q^{n-r}$

$\therefore$ r=4, i.e., x=4, 5, 6, 7

$\therefore \ P(X=4, 5, 6, 7)= \\ ^{7}C_{4} \left( \cfrac{1}{3} \right)^{4} \left ( \cfrac{2}{3} \right)^{3}+ ^{7}C_{5} \left( \cfrac{1}{3} \right)^{5} \left ( \cfrac{2}{3} \right)^{2}+^{7}C_{6} \left( \cfrac{1}{3} \right)^{6} \left ( \cfrac{2}{3} \right)^{1} +^{7}C_{7} \left( \cfrac{1}{3} \right)^{7} \left ( \cfrac{2}{3} \right)^{0}$

$\therefore \ P(X=4, 5, 6, 7)=0.1280 + 0.038 + (6.4014 \times 10^{-3}) + (4.5724 \times 10^{-4}) = 0.1728$

$\therefore$The expected number of times of getting (3 or 5) at least 4 times$=N \times p = 729 \times 0.1728=126.33 \approx 127$