nancyjose wrote:

Any explanation on the above? Thanks

The total number of outcomes = 2 ^5 =32 (it is because each toss has two possibilities Head or Tail.In general when a coin is tossed n times , the total number of possible outcomes = 2^n)

Let E = event of getting atleast 3 heads .

Atleast means that’s the minimum number of heads and it can be more than that ..ie he can get 3 heads , 4 heads or 5 heads.

So Favorable outcomes E ={3 heads and remaining 2 tails , 4 heads and 1 tail , 5 heads no tail}

={HHHTT,TTHHH,THHHT,HHHHT,THHHH,HTHHH,HHHTH, HHHHH}

So no. of favorable outcomes E =8

So the required probability = 8/32 = ¼