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Question: A box contains 6 red, 5 blue and 4 white marbles. Four marbles are chosen at random without replacem...

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?

Explanation

Solution

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(s)=6+5+4=15n\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(s)=15C4 ⇒n(s)=(15!)(11!)(4!) ⇒n(s)=1365  \,\,\,\,\,\,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(E)=P(RRBW  or  BBRW  or  WWRB) ⇒n(E)=6C2⋅5C1⋅4C1+5C2⋅6C1⋅4C1+4C2⋅6C1⋅5C1 ⇒n(E)=15⋅5⋅4+10⋅6⋅4+6⋅6⋅5 ⇒n(E)=300+240+180 ⇒n(E)=720  \,\,\,\,\,\,\,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:

\,\,\,\,\,\,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.