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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?

THE ANSWER IS NOT 7/16

Guest 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 \(\frac{16}{32}=\boxed{\frac{1}{2}}\).

lolzforlife  Jul 5, 2018
 #2
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That's correct.

Guest Jul 5, 2018
 #3
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No the answer is 13/16

Guest 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.

geno3141  Jul 5, 2018

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