Question

A box contains 5 different red and 6 different white balls. In how many ways can 6 balls be taken out so that there are at least two balls of each color?

Answer 1

Hint: Event is a subset of the sample space, i.e. a set of outcomes of the random experiment.
\[{\text{Probability of an event}} = \dfrac{{{\text{Number of occurence of event A in S}}}}{{{\text{Total number of cases in S}}}} = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}{\text{ }}\]
In this question, there is the use of the concept of combination and probability simultaneously to get the number of ways in which 6 balls can be taken out so that there are at least two balls of each color.

Complete step-by-step answer:
A box contains 5 different red balls and 6 different white balls. So, the number of sample space of drawing 6 balls be taken out so of 11 balls is given as:
 $
\Rightarrow n(S) = {}^{11}{C_6} \\
   = \dfrac{{11!}}{{6!5!}} \\
   = \dfrac{{11 \times 10 \times 9 \times 8 \times 7 \times 6!}}{{6!{\text{ }} \times 5 \times 4 \times 3 \times 2 \times 1}} \\
   = 11 \times 6 \times 7 \\
   = 11 \times 42 \\
   = 462 \\
  $
Now, the favorable cases are drawing at least two balls of one color is:
 $
\Rightarrow n\left( E \right) = \left( {2{\text{ }}red,{\text{ }}4{\text{ }}white} \right) + \left( {3{\text{ }}red,{\text{ }}3{\text{ }}white} \right) + \left( {4{\text{ }}red,{\text{ }}2{\text{ }}white} \right) \\
\Rightarrow n(E) = {}^5{C_2}.{}^6{C_4} + {}^5{C_3}.{}^6{C_3} + {}^5{C_4}.{}^6{C_2} \\
\Rightarrow n(E) = 10 \times 15 + 10 \times 20 + 5 \times 15 \\
\Rightarrow n(E) = 150 + 200 + 75 \\
\Rightarrow n(E) = 425 \\
  $
Hence, the required probability is:
 $ \dfrac{{n(E)}}{{n(S)}} = \left( {\dfrac{{425}}{{462}}} \right) $

Additional Information: The combination is a way of selecting items from a collection, such that the order of selection does not matter. The number of the combination (selection) of r things out of n different things without replacement and where order does not matter is given as: $ {}^n{C_r} = \dfrac{{n!}}{{{\text{ r!}}\left( {n - r} \right)!}} $
The number of combinations of n different items taken nothing at all is considered as 1. Counting combinations is counting the number of ways in which all or some of the objects are selected at a time. Therefore, there is only one way of not selecting an item.

Note: n case of at least 2 red balls, in the solution, the cases of 1 red and 0 red ball are not taken. A similar concept is used for white balls. There are three different cases.