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Coin A is tossed three times and coin B is tossed two times. What is the probability that more heads are tossed using coin A than coin B?

Jul 5, 2018

#1
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Coin $$A$$can either have more heads or less heads than Coin $$B$$, but not both.

The total number of possibilities for the coin flips is $$8\cdot 4=32$$.

Subtracting the number of flips that have less heads for coin $$A$$or an equal amount of heads makes $$\frac{16}{32}=\boxed{\frac{1}{2}}$$.

Jul 5, 2018
#4
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Actually it's 1/2 because you have to subtract the possibilities where it is an equal amount of heads.  That was specified in the answer, so...$$\color{gold}{\colorbox{blue}{you're wrong.}}$$

Guest Jul 5, 2018
#5
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The possible ways to toss coin B:                 The possible ways to toss coin A:

H H                                                                          H H H                             H T T

H T                                                                           H H T                             T H T

T H                                                                           H T H                             T T H

T T                                                                           T H H                              T T T

The probability of getting two Heads with coin B and three Heads with coin A:            (1/4)(1/8)  =  1/32.

The probability of getting one Head with coin B and two or three Heads with coin A:  (2/4)(4/8)  =  8/32.

The probability of getting no Heads with coin B and at least one Head with coin A:     (1/4)(7/8)  =  7/32.

Adding together these possibilities:  1/32  +  8/32  +  7/32  =  16/32  =  1/2.

Jul 5, 2018