Three numbers are chosen at random without replacement from {1, 2, 3, …,
10}. What is the probability that the minimum of the chosen number is 3, or their
maximum is 7?
Approach 1:
Let A and B denote the following events
A:minimum of the chosen number is 3
B: maximum of the chosen number is 7
We have,
P(A)=P(choosing 3 and two other numbers from 4 to 10)=${{7\choose2}\over{10\choose3}}={(7*6)\over2}*{3*2\over10*9*8}={7\over40}$
P(B)=P(choosing 7 and two other numbers from 1 to 6)=${{6\choose2}\over{10\choose3}}={6*5\over2}*{3*2\over10*9*8}={1\over8}$
P(A∩B)=P(Choosing 3 and 7 and one other number from 4 to 6)$={3\over{10\choose3}}={3*3*2\over10*9*8}={1\over40}$
Now P(A∪B)=P(A)+P(B)−P(A∩B)
$={7\over40}+{1\over8}-{1\over40}={11\over40}$
Approach 2:
Chance of a number being drawn out of the given 10 numbers is equiprobable. Hence
P(any one number drawn )=${1\over10}$
P(A)=${7\choose2}{1\over10}*{7\over9}*{6\over8}=21*{42\over720}={14*7\over80}={98\over80}$ which is wrong as it is above 1.
Here, ${1\over10}$ is the probability of getting 3 and ${7\over9}*{6\over8}$ is the probability of getting two number above 3 which can be chosen in ${7\choose2}$ ways.
**Why I cannot find the P(A) like this ?
Please tell me what is wrong in approach 2 **
Best Answer
Your approach 2 has three errors. First, you are double multiplying by $7 \choose 2$, once when you use the binomial and once in the numerators of $\frac 79$ and $\frac 68$. When you use $\frac 79$ and $\frac 68$ you are already choosing the numbers. You have a factor $2$ when you use $\frac 79$ and $\frac 68$ (note $7 \cdot 6=2{7 \choose 2})$ because there are two orders to select each pair and you count them both. Finally, you only count cases where $3$ is the first number selected.