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

compute the probability that none land head, subtract the result from 1
since "at least one" means "not none"

Answer 2

ok i lies, it says more than one" so 2 or 3 or 4 or 5

Answer 3

compute the probability you get none or one
then subtract those from 1

Answer 4

\[\large 1-\left(\frac{1}{2}\right)^5-5\times \left(\frac{1}{2}\right)^5\]

Answer 5

@satellite73 dont mean to bother you but are you done helping me out?

Answer 6

@zepdrix hey man, are you available to give me some help?

Answer 7

@jim_thompson5910 hey jim, do you know anything about "two's compliment?"

Answer 8

what's your question?

Answer 9

Ok, just a moment

Answer 10

1. 0001110001110001
2. 1110001110001110
+ 1
1110001110001111

Answer 11

oh, wait a minute. i think i just saw what i did wrong... i wasn't supposed to flip it before converting it because it's positive already

Answer 12

have another question!

Answer 13

Earlier in this thread Satelite gave me an equation that solves the original post. I don't understand what the parts of it mean 1-(1/2)^5 - 5x(1/2)^5

Answer 14

I'm guessing that (1/2) is the probability for either side. ^5 is because we have 5 flips. but, why the 1 and the 5? if i wanted to say "more than two flips would come up heads" which number would i change?

Answer 15

@jim_thompson5910

Answer 16

one moment, still thinking

Answer 17

The probability of getting heads is 1/2

The probability of getting all five heads is (1/2)^5

the complement of this event is having at least one tail, so that would be 1 - (1/2)^5

Answer 18

you mean that 5(1/2)^5 is the probability of 5 heads?

Answer 19

I mean (1/2)^5 is the probability of getting 5 heads in a row

Answer 20

okay, i can see that now. but, why does the -5 come in?

Answer 21

not sure but I'm still thinking

Answer 22

I think it has to do with the probability of getting exactly one tail

Answer 23

this is important concept for me to grasp tonight cause this is one of the chapters on my test tomorrow :(

Answer 24

I'm guessing you're being tested on the binomial probability theorem?

Answer 25

Yes, i think so. I just looked through my book, though and i don't see that exact term.

Answer 26

have you seen this formula before

\[\Large \left(_n C _k\right)*p^k*(1-p)^{n-k}\]

Answer 27

no, but i'm trying to find it

Answer 28

is that what satelite used?

Answer 29

not quite

Answer 30

what I posted is the binomial probability formula. It's used to answer questions like "what is the probability of getting exactly one tail?"

Answer 31

there are 4 variables in your formula :(

Answer 32

How would I plug my problem into this monster?

Answer 33

n = 5 because there are 5 quarters (5 trials)
p = 0.5 is the probability of getting tails
k = 1 if we are asking "what is the probability of getting exactly one tail?"

Answer 34

C is NOT a variable. It stands for the combination formula

\[\Large _nC_r = \frac{n!}{r!(n-r)!}\]

Answer 35

oh!

Answer 36

I know that notation

Answer 37

you can use a calculator to compute the combination or you can use Pascal's Triangle

Answer 38

Wow, so what Satelite did was wrong?

Answer 39

that seems like a lot of math

Answer 40

P(more than 1 tail) = 1 - P(no tails) - P(exactly one tail)

we already calculated `P(no tails)` when we computed (1/2)^5. We could have used the binomial theorem but no need here

you just need `P(exactly one tail)` now

Answer 41

what would r be?

Answer 42

@satellite73 hey man, could you explain the formula you gave me earlier to answer this question?

Answer 43

in the case of exactly 1 tail, r = 1