n coins are flipped, and you are given that at least 1 of them is heads. The probability that at least one other coin is heads is 19/21. How many coins do you have?
Hello, mark1950!
An unusual problem . . .
I had to come up with an unusual solution.
When $\displaystyle n$ coins are flipped, there are $\displaystyle 2^n$ possible outcomes.$\displaystyle n$ coins are flipped, and you are given that at least 1 of them is heads.
The probability that at least one other coin is heads is $\displaystyle \frac{19}{21}$
How many coins do you have?
We are told that at least one of the coins was Heads.
. . Then there are: .$\displaystyle 2^n - 1$ outcomes.
We eliminate the case of "no Heads" (all Tails).
We want the probability that at least two coins are Heads.
. . There is 1 outcomes with 0 Heads.
. . There are $\displaystyle n$ outcomes with exactly 1 Head.
Hence, there are: .$\displaystyle 2^n - n - 1$ outcomes with at least 2 Heads.
The probability of at least two Heads, given there is at least one Head is $\displaystyle \frac{19}{21}$
. . $\displaystyle \frac{2^n - n - 1}{2^n - 1} \;=\;\frac{19}{21} \quad\Rightarrow\quad 21\cdot2^n - 21n - 21 \;=\;19\cdot2^n - 19 $
. . $\displaystyle 2\cdot2^n - 21n \:=\:2 \quad\Rightarrow\quad 2^{n+1} \;=\;21n + 2$
There is no elementary method for solving this equation.
By inspection, I found the solution: .$\displaystyle n = 6$
Erm, I don't really get this step. If you don't mind, can you please explain in a more specific detail? Thanks.
We want the probability that at least two coins are Heads.
. . There is 1 outcomes with 0 Heads.
. . There are n outcomes with exactly 1 Head.
Hence, there are: .2^n - n - 1 outcomes with at least 2 Heads.
And also, how do you do inspections to get n = 6?
If Soroban doesn't mind my jumping in here:
With n coins there are $\displaystyle 2^n$ possible outcomes: each coin can come up either heads or tails so each coin multiplies the possiblities by 1. Exactly one of those has "all tails" so $\displaystyle 2^n-1$ have "at least one head.
There are exactly n outcomes with exactly one head because there are n positions that "head" can be in. To find the number of possible outcomes with 2 or more heads, start with $\displaystyle 2^n$ total and subtract off the number with one head or no heads: $\displaystyle 2^n- n- 1$. The probability that "given that there is at least one head, there are at least 2" is $\displaystyle \frac{2^n-n-1}{2^n-1}$ and that must be 19/21.
Now just calculate a few values: if n= 2 (there must be at least two flips to get two heads!), $\displaystyle \frac{2^2- 2-1}{2^n-1}= \frac{1}{3}$.
If n= 3, $\displaystyle \frac{2^3- 3- 1}{2^3-1}= \frac{4}{7}$.
If n= 4, $\displaystyle \frac{2^4- 4- 1}{2^4-1}= \frac{11}{15}$
If n= 5, $\displaystyle \frac{2^5- 5- 1}{2^5-1}= \frac{26}{31}$
if n= 6, $\displaystyle \frac{2^6- 6- 1}{2^6-1}= \frac{57}{63}= \frac{19}{21}$!