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A penny is tossed 3 times and a nickel is tossed 2 times. What is the probability that more heads are tossed using the penny than using the nickle?

 May 6, 2021
 #1
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Well, we need to determine how many outcomes of flipping would result in a penny "win" vs a penny "tie or loss".

 

For a penny, there's 2^3 possibilities (number of options to the power of number of flips), and for nickels it'd be 2^2. The chance of getting each result is then "x in 4" or "x in 8".

 

Outcome Penny Nickel
0 heads 1 in 8 1 in 4
1 head 3 in 8 2 in 4
2 heads 3 in 8 1 in 4
3 heads 1 in 8 0 in 4

 

So add the results of the penny runs that beat the each nickel run together:

If Nickels get zero, Pennies win 7 in 8 runs

If Nickels get one, Pennies win 4 in 8 runs

if Nickels get two, pennies win 1 in 8 runs

 

Multiply each of these by its weight (i.e. how likely the scenario is to occur):

7 in 8 * 1 in 4 ||\(7/8 * 1/4 = 7/32\)

4 in 8 * 2 in 4 || \(4/8 * 2/4 = 1/4 or 8/32\)

1 in 8 * 1 in 4 || \(1/8 * 1/4 = 1/32\)

So I see 16 in 32 or simply 1 in 2.

 

Half the time, Penny wins :)

 May 6, 2021
 #2
avatar+2209 
+1

It seems a bit weird that nickel has equal heads as pennies. 

I think you forgot the scenerio where nickel has 3 heads. 

 

=^._.^=

catmg  May 6, 2021
edited by catmg  May 6, 2021
 #3
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It seems pretty appropriate. It's harder for the penny to win but it also gets an additional flip. If you kindly refer back to the question, the asker specifies 3 flips for penny and 2 flips for nickels.

Guest May 7, 2021
 #4
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0

Oops, my bad. 

My brain stopped working for a moment.

Thank you for correcting me. :))

 

=^._.^=

catmg  May 7, 2021

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