Permutations and Combinations Part 1

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
=

33

81

×27
=11