If you toss 1000 fair coins 10 times each what is the probability that *some* coin will get 10 heads?
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.
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.
Hello, mathquest!
My solution is a rehash of Milokerr90's solution . . .
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: .$\displaystyle \left(\frac{1}{2}\right)^{10} \:=\:\frac{1}{1024}$
The probability of a coin not getting 10 Heads: .$\displaystyle 1 - \frac{1}{1024} \:=\:\frac{1023}{1024}$
The probability of 1000 coins, not getting 10 Heads: .$\displaystyle \left(\frac{1023}{1024}\right)^{1000}$
The probability of 1000 coins, some getting 10 Heads:
. . . . $\displaystyle 1 - \left(\frac{1023}{1024}\right)^{1000} \;=\;0.623576202 \;\approx\;62.3\%$