Coin A is flipped 3 times and Coin B is flipped 2 times, what is the probability that Coin B has more heads than Coin A?

$\frac5{32}$
$\frac3{16}$
$\frac18$
$\frac7{32}$

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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

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

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

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