# combinatorics and coins

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• Apr 8th 2012, 03:09 PM
mathquest
combinatorics and coins
If you toss 1000 fair coins 10 times each what is the probability that *some* coin will get 10 heads?
• Apr 8th 2012, 03:15 PM
mathquest
Re: combinatorics and coins
answer: aproximatly 63%
and my question is ..why?

thnx!
• Apr 8th 2012, 03:16 PM
Milokerr90
Re: combinatorics and coins
The use of *some* coin getting 10 heads suggests that you are interested in the probability that at least one coin gets ten heads.

If we let N=number of coins getting 10 heads, then Pr(N>=1)=1-Pr(N=0). Therefore, if we can find the probability that none of the 1000 coins get 10 heads, we can calculate your needed answer.
• Apr 8th 2012, 03:18 PM
mathquest
Re: combinatorics and coins
I can't see if this problem is easier to get solved.
• Apr 8th 2012, 03:29 PM
Milokerr90
Re: combinatorics and coins
If we calculate the probability that one of the 1000 coins does not get 10 heads, then we raise this to the power of 1000 to get the probability that none of the 1000 coins get 10 heads as the events are independent.

So, one way to find this would be to find the probabilities of getting 1 head, 2 heads,...,9 heads out of 10 and adding these together:

(1/2)+(1/2)^2+...+(1/2)^9=0.999023438

Or, we could calculate 1-(Probability of getting 10 heads)=1-(1/2)^10=0.999023438. Now, we raise this to the power of 1000 as we have 1000 coins, and find the probability that none of the 1000 coins get 10 heads=0.999023438^1000=0.376423986, then take this away from 1 to find the probability that at least one gets 10

1-0.376423986=0.623=62.3%

This is one of my first replies so please, feel free to ask questions if this isn't clear.
• Apr 8th 2012, 03:46 PM
mathquest
Re: combinatorics and coins
It's very clear solution and explanation! Thank you :)
• Apr 8th 2012, 04:45 PM
Soroban
Re: combinatorics and coins
Hello, mathquest!

My solution is a rehash of Milokerr90's solution . . .

Quote:

If you toss 1000 fair coins 10 times each,
. . what is the probability that some coin will get 10 heads?

The probability of one coin getting 10 Heads: . $\left(\frac{1}{2}\right)^{10} \:=\:\frac{1}{1024}$

The probability of a coin not getting 10 Heads: . $1 - \frac{1}{1024} \:=\:\frac{1023}{1024}$

The probability of 1000 coins, not getting 10 Heads: . $\left(\frac{1023}{1024}\right)^{1000}$

The probability of 1000 coins, some getting 10 Heads:

. . . . $1 - \left(\frac{1023}{1024}\right)^{1000} \;=\;0.623576202 \;\approx\;62.3\%$

• Apr 20th 2012, 07:54 AM
mathquest
Re: combinatorics and coins
Thank you Soroban!
Good rehash!!!