Step 1

Given:

An urn that contains 12R, 16B, 18G balls

7 balls are randomly chosen

Probability is equally distributed over the outcome space S:

\(S=\{\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}\} ; x_{i}\) are different balls from the urn

Since S contains 7 valued combinations \(\displaystyle{\left|{S}\right|}=\)

\(\binom{46}{7}\)

(\(\displaystyle{\left|{X}\right|}\) is the number of elements in X)

For an event \(A\subseteq S\), \(\displaystyle{\left|{A}\right|}={n}\)

\(P(A)=\frac{n}{\binom{46}{7}}\)

a) \(\displaystyle{P}{\left({3}{R},{2}{B},{2}{G}\right)}=?\)

Find the number of 7 valued combinations of balls where three are red, two blue and 2 green.

Choose the three red (R) from 12 of them in \(\binom{12}{3}\) ways, then choose 2 more blue to generate \(\binom{16}{2}\) times more possibilities, and choose the 2 green ones - \(\binom{18}{2}\) possibilities

\(P(3R, 2B, 2G)=\frac{\binom{12}{3}\cdot\binom{16}{2}\cdot \binom{18}{2}}{\binom{46}{7}}=\frac{3060}{40549}=0.075464\)