Answer:
![\text{Probability of atleast two heads is}=\frac{11}{16}](https://tex.z-dn.net/?f=%5Ctext%7BProbability%20of%20atleast%20two%20heads%20is%7D%3D%5Cfrac%7B11%7D%7B16%7D)
Step-by-step explanation:
We have total of 16 possible outcomes for flipping of four coins
We need to find the probability when atleast two head comes that means it should be atleast 2 head means it can be more than 2 so, can be 3 or 4 also
We will consider the cases of 2,3,4 heads on coins
So, the cases of 2 heads are HHTT ,HTHT ,HTTH, THHT ,THTH ,TTHH =6 outcomes
The case of 3 heads are HHHT ,HHTH ,HTHH, THHH=4 outcomes
The case of 4 heads are HHHH=1 outcome
So, the probability of atleast two head will be 6+4+1=11
![\text{Probability of atleast two heads is}=\frac{11}{16}](https://tex.z-dn.net/?f=%5Ctext%7BProbability%20of%20atleast%20two%20heads%20is%7D%3D%5Cfrac%7B11%7D%7B16%7D)