Since the problem states "at least" we need to also find probability of 3 H or 4 H or 5 H
Now find the probability of flipping a head 4 times; [tex] \frac{1}{2} [/tex]⁴
= (1/16)
Now probability of flipping a head 3 times: (4C3)(1/2)⁴ = 4/16
Probability of flipping a head 2 times; (4C2)(1/2)⁴=6/16
(1/16)+(4/16)+(6/16)=11/16
Probability of flipping a fair coin 4 times with at least 2 heads is 11/16.
Hope I helped :)
