3 vs 2

Let $A$ be the number of heads resulting from coin $A$ being thrown three times, and $B$ be the number of heads resulting from coin $B$ being thrown twice. Then since the outcomes for coin $A$ and coin $B$ are independent, we have that

$P(A \lt B) = P(A=0, B=1) + P(A=0, B=2) + P(A=1, B=2) =$

$P(A=0)*P(B=1) + P(A=0)*P(B=2) + P(A=1)*P(B=2) =$

$\dfrac{1}{8}*\dfrac{1}{2} + \dfrac{1}{8}*\dfrac{1}{4} + \dfrac{3}{8}*\dfrac{1}{4} = \boxed{\dfrac{3}{16}}.$

(probability A got 0 head) x (probability B got 1 or 2 heads) + (probability A got 1 head) x (probability B got 2 heads)

= (1/8 x 3/4) + (3/8 x 1/4)

= 3/32 + 3/32

= 6/32

= 3/16

= 18.75 %

assumptions: probability of head=0.5 tails=0.5

coin b--> Head,Head=0.5^2 coin A must have 0 or 1 head so... head,tails,tails tails,head,tails tails,tails,head so 3 multiplied by 0.5^3. (0.5^2)(0.5^3)(3)=3/32

0 heads= tails tails tails=0.5^3 if coin B has 1 head, Coin A must have zero heads. tails, tails,tails (0.5^3)(0.5^2)=(1/32)

again, Coin b has one head. so tails,head (probability=0.5^2) coin a=Tails tails tails (1/32 once again)

Other way round, tails then head (same probability of 0.5^2) tails,tails,tails (0.5^2)(0.5^3)=(1/32)

(3/32)+3(1/32)=3/16