Question

A child tosses five quarters in the air. Each quarter lands either falls so that “heads” is visible or “tails” is visible. What is the probability that more than one quarter lands on tails?

Answer 1

@satellite73 I re-posted it. You gave 1-(1/2)^5 x 5-(1/2)^5. I am not sure how the 1 and 5 are fitting into this puzzle here.

Answer 2

@mathmate @zepdrix @kropot72 if anyone else wants to walk me through this process that would be awesome.

Answer 3

the probability that all land heads is \[(\frac{1}{2})^5\] or \[\frac{1}{2^5}\]is that clear?

Answer 4

Yes

Answer 5

I just don't understand how the values 1 and 5 are fitting into this

Answer 6

so how about the probability that exactly one tail and four heads?

Answer 7

any particular sequence has probability \[\frac{1}{2^5}\]

Answer 8

there are five such sequences with 1 tail and two heads
t h h h h
h t h h h
h h t h h
h h h t h
h h h h t

Answer 9

each has the same probability and they are all different, so you add them up i.e. multiply by 5 to get \[\frac{5}{2^5}\]

Answer 10

Okay. It was actually - 5(1/2)^5 at the end (just a quick correction).

Answer 11

ok now lets go slow

Answer 12

you want "more than one quarter lands on tails?"

Answer 13

more than one means 2 or 3, or 4, or 5
you can compute each of those and add to get your answer, but there is a shorter way

Answer 14

Yes.

Answer 15

but it is shorter to compute the probability you get 0 or 1

Answer 16

that is what is left
so compute those two, add them together, subtract the result from one
that way you only have to compute two probabilities instead of 4

Answer 17

i am going to try just rereading over this to see if i can't get it without wasting your time

Answer 18

ok ask if you have any questions
i can give another example quickly if you likke

Answer 19

heh

Answer 20

im just terrible at math, dude

Answer 21

I can see that 5(.03125)^5 is each combination where there is 1 heads

Answer 22

but i dont understand the rest of it.

Answer 23

The probability of zero tails on all 5 coins is (1/2)^5.
Using the binomial distribution, the probability of exactly one tail among the 5 coins is given by:
\[\large P(1\ tail)=5C1\times(\frac{1}{2})^{1}\times(\frac{1}{2})^{4}=\frac{5}{2^{5}}\]
Adding the probabilities of zero tails and one tail we get:
\[\large P(0\ or\ 1\ tail)=\frac{6}{2^{5}}\]
So the probability of more than one tail is
\[\large P(more\ than\ 1\ tail)=1-\frac{6}{2^{5}}=you\ can\ calculate\]

Answer 24

:o

Answer 25

yeah

Answer 26

Make sense of all this yet? :)

Answer 27

no man. this last guy didn't even go 5(1/2)^5

Answer 28

Hmm I'm not sure if my explanation is going to be any better,
I would have done it very similar to Kropot's way.

Answer 29

i feel so dumb

Answer 30

We have 5 slots, two options for each slot.
So `2^5 total` possible combinations of flips.

How many ways are there to flip `0 tails`?
1 way, all heads.

how many ways are there to flip `1 tails`?
5 different ways.

So we can flip 1 tail or less in 6 ways.
Then we can flip more than 1 tail in `Total` - `6` ways.

These are the total number of flips which result in more than 1 tails:
\(\large\rm 2^5-6\)

But they want the probability, so we need to divide this amount by the total number of possible flips, our sample space.

\(\large\rm \dfrac{2^5-6}{2^5}\)
Doing some division gives us:
\(\large\rm 1-\frac{6}{2^5}\)

Answer 31

Do you have answer key for this problem? :U

Answer 32

yeah

Answer 33

how did you pull 1 out of that :X?

Answer 34

\[\large\rm \frac{2^5-6}{2^5}\quad=\quad \frac{2^5}{2^5}-\frac{6}{2^5}\]

Answer 35

oh, duh, sorry.

Answer 36

oh, your explanation is better. i can smell the synapses firing!

Answer 37

or burning. not sure

Answer 38

lol XD

Answer 39

I find probability stuff to be incredibly confusing.
It never made sense to me like other math does.
So I feel yer pain :U

Answer 40

i just have a hard time understanding what is going on inside the formulas with this

Answer 41

ok, so i'm going to try to figure out what it would be for 3 tails.

Answer 42

For exactly 3 tails?

Answer 43

yeah

Answer 44

how would i do that without just counting everything up? :(

Answer 45

I think it's just \(\left(\begin{matrix}5 \\ 3\end{matrix}\right)\)

Thinking...

Answer 46

what is that c(5,3)?

Answer 47

Nah that can't be right :) sec thinking XD

Answer 48

yes, combination

Answer 49

Answer key? XD

Answer 50

that's a problem i just made up

Answer 51

ah :)

Answer 52

https://www.youtube.com/watch?v=udG9KhNMKJw

Answer 53

i want to make sure i can use this for flipping coins when it comes up on the test tomorrow.. which i'm probably screwed on

Answer 54

ok i'm watching link

Answer 55

Ah yes it is simply 5 choose 3, it worked yay!

Answer 56

We have 5 places to put 3 tails,
order does not matter,
so \(\large\rm C(5,3)=\frac{5!}{3!(5-3)!}=~...~=10\)

Answer 57

well i'll be damned

Answer 58

so if i wanted the probability of flipping > 2heads i could calculate it that way

Answer 59

But again, if you wanted a `probability`,
the odds of getting exactly 3 tails would be: \(\large\rm \dfrac{10}{2^5}\)

Answer 60

Ummm

Answer 61

it's not a probability?

Answer 62

it's out of 1....

Answer 63

Hmm? :o

Answer 64

1 = 100%

10/32 = 31.25%

So the odds of rolling exactly 3 tails is about 31 percent chance.

Answer 65

no no no

Answer 66

you've got it all wrong.

Answer 67

your odds of (F)lipping 3 tails is 31%

Answer 68

Ya sorry I was thinking of dice I guess :) lol rolling

Answer 69

can't we just stick a p infront of it and call it a day? p(31%)

Answer 70

bam!

Answer 71

;p

Answer 72

But to your other question, about using this shortcut for >2,

Yes, that should work out fine!
C(5,3) for all the combinations involving 3 tails,
C(5,4) for all the combinations involving 4 tails,
C(5,5) for all the combinations involving 5 tails.

So all the combinations involving more than 2 tails flipped is
C(5,3) + C(5,4) + C(5,5)

Then the probability of flipping more than 2 tails is\[\large\rm \frac{C(5,3)+C(5,4)+C(5,5)}{2^5}\]

Answer 73

yeah... did we already go over how to make this into a probability? im going to scroll up and check

Answer 74

ummm

Answer 75

so we kinda did just stick a p infront of it almost

Answer 76

You divide your
`number of possibilities in the given event`
by the
`total number of possibilities in the sample space`

We determined earlier that the size of the space is 2^5,
that's the total number of possible outcomes from flipping 5 coins.

What we were doing separately was counting the number of outcomes when more than 2 tails are flipped.

Then we divide that by the total for our % in relation to the total.

Sorry I'm not really familiar with the P notation :\