4 dice are independently thrown. Each die has probability to show 3 or 5 is p=
2
6
=
1
3
∴q=1−
1
3
=
2
3
(not showing 3 or 5) Experiment is performed with 4 dices independently. ∴ Their binomial distribution is (q+p)4=(q)4+‌4C1q3p+‌4C2q2p2+‌4C3qp3+‌4C4p4 ∴ In one throw of each dice probability of showing 3 or 5 at least twice is =p4+‌4C3qp3+‌4C2q2p2 =
33
81
∴ Such experiment performed 27 times ∴ so expected out comes =np =