<s>[INST] Probability problem gives different answers with different methods 
In a box there are $4$ red ball, $5$ blue balls and $6$ green balls.$4$ balls are picked at random. What is the probability the at least one ball is green and at least one ball red $?$

Method $1$:
Now,in how many ways can this be done $?$
$1.$ $1$ red,$1$ green,$2$ blue.
$2.$ $2$ red,$1$ green,$1$ blue.
$3.$ $1$ red,$2$ green,$1$ blue. 
$4.$ $2$ red,$2$ green,$0$ blue.
$5.$ $3$ red,$1$ green,$0$ blue.
$6.$ $1$ red,$3$ green,$0$ blue.
So the all possible cases are obtained by adding these cases,i.e. , 
$${^6C_1} \times {^5C_2} \times {^4C_1} + {^6C_2} \times {^5C_1} \times {^4C_1} + {^6C_1} \times {^5C_1} \times {^4C_2} + {^6C_2} \times  {^4C_2} + {^6C_3} \times {^4C_1} + {^6C_1}  \times {^4C_3} \\= 914.$$
So the probability is ${914\over {({_{15} C_4})}}={914\over 1365}$
Method $2$:
Probability of at least $1$ red ball and at least $1$ green ball = $1$- Probability of no red or green ball, i.e. all balls are blue =$1-{5\over 1365} = {1360\over 1365}.$
Why these two methods give different answers? Are one or both of them wrong $?$ How?
 [/INST] Let $R$ be the number of reds and $G$ be the number of greens. 
Then
\begin{align*}
P(R\geq 1, G\geq 1) &= 1-P(R=0 \cup G = 0) \\
&= 1-[P(R=0)+P(G=0)-P(R=0, G=0)]\\
&= 1 -\left[\frac{\binom{4}{0}\binom{11}{4}}{\binom{15}{4}}+\frac{\binom{6}{0}\binom{9}{4}}{\binom{15}{4}}-\frac{\binom{5}{4}\binom{10}{0}}{\binom{15}{4}}\right] \\&= \frac{914}{1365}\end{align*}
where the second line is true by inclusion-exclusion. This agrees with your first answer.
</s>