Question

A box contains 6 red, 5 blue and 4 white marbles. Four marbles are chosen at random without replacement. The probability that there is at least one marble of each color among the four chosen, is?

Answer 1

Hint: In this problem, first find the total number of marbles. There are 3 cases for the selection of 4 marbles so that at least one marble of each color is chosen. Next, find the probability by dividing the favorable outcome by sample space.

Complete step by step solution:
Total number of marbles \[n\left( s \right) = 6 + 5 + 4 = 15\].
The total number of red marbles is 6.
The total number of blue marbles is 5.
The total number of white marbles is 4.
Probability of selecting 4 things among 15 is shown below.
\[
  \,\,\,\,\,\,n\left( s \right) = {}^{15}{C_4} \\
   \Rightarrow n\left( s \right) = \dfrac{{\left( {15!} \right)}}{{\left( {11!} \right)\left( {4!} \right)}} \\
   \Rightarrow n\left( s \right) = 1365 \\
\]
Total number of favorable outcomes is calculated as follows:
\[
  \,\,\,\,\,\,\,P\left( E \right) = P\left( {{\text{RRBW}}\,\,{\text{or}}\,\,{\text{BBRW}}\,\,{\text{or}}\,\,{\text{WWRB}}} \right) \\
   \Rightarrow n\left( E \right) = {}^6{C_2} \cdot {}^5{C_1} \cdot {}^4{C_1} + {}^5{C_2} \cdot {}^6{C_1} \cdot {}^4{C_1} + {}^4{C_2} \cdot {}^6{C_1} \cdot {}^5{C_1} \\
\Rightarrow n\left( E \right) = 15 \cdot 5 \cdot 4 + 10 \cdot 6 \cdot 4 + 6 \cdot 6 \cdot 5 \\
\Rightarrow n\left( E \right) = 300 + 240 + 180 \\
\Rightarrow n\left( E \right) = 720 \\
\]
The probability that there is at least one marble of each color among the four chosen is calculated as follows:
\[\begin{gathered}
  \,\,\,\,\,\,P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}} \\
   \Rightarrow P\left( E \right) = \dfrac{{720}}{{1365}} \\
   \Rightarrow P\left( E \right) = \dfrac{{48}}{{91}} \\
\end{gathered}\]
Thus, the probability that there is at least one marble of each color among the four chosen marbles is \[\dfrac{{48}}{{91}}\].

Note: The formula for the section of r items among n items is shown below.
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Here, the symbol “!” represents factorial.