# 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?

• Method 2 is wrong because you also have to minus the probability that there is green and blue balls, or red and blue balls, but not having red and green balls at the same time. – Lazy Lee May 18 '17 at 4:53
• "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". This is false - suppose you get 1 red and 3 blues. Then you do have at least 1 red, but you don't have the "at least 1 green$which is necessary too. – John Doe May 18 '17 at 4:54 ## 1 Answer Let$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.