A 7 digit number is formed by arranging the digits 1,2,2,3,6,6,6. What is the probability the number formed is larger than 3,000,000?
Can someone tell me if my approach makes sense/and or is correct.
So in order to be greater than 3,000,000 the first digit needs to be a 3,6,6,6 so there are 4 options for the first spot and then count down for the rest so 4x6x5x4x3x2x1 then for all total arrangements that would be 7! so the probability is 2880/5040. Is that correct?
On
The total number of numbers equals:$$\frac{7!}{1!2!1!3!}$$
The number of numbers that have $6$ as first digit equals:$$\frac{6!}{1!2!1!2!}$$
The number of numbers that have $3$ as first digit equals:$$\frac{6!}{1!2!0!3!}$$
If every number has equal probability to occur then:$$\left(\frac{6!}{1!2!1!2!}+\frac{6!}{1!2!0!3!}\right)/\frac{7!}{1!2!1!3!}=\frac47$$ is the probability that a number is formed that starts with $3$ or $6$, i.e. exceeds $3000000$.
There is a shorter route to this result.
If one of the numbers is randomly chosen then the probability that its first digit will be a $6$ or a $3$ is $\frac47$ since exactly $4$ of the $7$ digits will belong to $\{3,6\}$.
You would be correct if the digits were painted colors and you thought a number with a blue $6$ in the millions place was different from a number with a red $6$. As they are not painted, you need to divide by the number of ways to rearrange the $2$s and $6$s to get the same number. If we were just asked the number of numbers we could form from $1,1,2$ it is only $3$, not $6$.
A simpler approach is to note that you get a number larger than $3,000,000$ any time the first digit is $3$ or $6$. That is four digits out of seven.