Question 1202771
It's a binomial with n=12 and p=0.7
the expected value is 8.4 That answers the first part.
For probability of 3 or more, do 1-prob (0,1,2)
prob of 0=0.3^12=0 essentially 
prob of 1 is 12*0.7*0.3^11=0.00001
prob of 2 is 12C2*0.7^2*0.3^10=0.0002
answer is the 1-sum which is 1-0.0002=0.9998
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can check the mode 12C8*0.7^8*0.3^4=0.2311
and for 9 is 12C9*0.7^9*0.3^3=0.2397
The mode is 9.