Question

In a bag, there are infinitely many red, white and black balls, which are identical. If ten balls are selected at random then the probability that selection includes at least one ball from each colour is:
A) \[\dfrac{5}{{11}}\]
B) \[\dfrac{6}{{11}}\]
C) \[\dfrac{7}{{11}}\]
D) \[\dfrac{4}{{11}}\]

Answer 1

Hint:
Here, we will use the probability of any event happening is given by dividing the number of outcomes of that event divided by the total number of events, that is; $P = \dfrac{{{\text{Number of outcomes}}}}{{{\text{Total number of outcomes}}}}$. Then we will take the total number of outcomes is 3. Now we will find the probability of getting the red ball from the three colours balls, probability of getting the white ball from the three colours balls and probability of getting the black balls from the three colours balls. Then we will assume that the probability for selection induces at least one ball from each colour to be found is \[{\text{P}}\]. So, we have \[{\text{Red}} + {\text{Black}} + {\text{White}} = 10\], such that \[{\text{Red}} \geqslant 1\], \[{\text{White}} \geqslant 1\] and \[{\text{Black}} \geqslant 1\]. Now we will find the total number of ways for selecting from the given conditions. Then we will divide the total ways for selecting by the ways such that no constraints on quantity.

Complete step by step solution:
We are given that in a bag there are infinitely many red, white and black balls, which are identical.
We know that the probability of any event happening is given by dividing the number of outcomes of that event divided by the total number of events, that is; $P = \dfrac{{{\text{Number of outcomes}}}}{{{\text{Total number of outcomes}}}}$.
Since we know that there three colours of balls, so the total number of outcomes is 3.
Now we will find the probability of getting the red ball from the three colours balls, we get
\[ \Rightarrow {P_{RED}} = \dfrac{1}{3}\]
Now we will find the probability of getting the white ball from the three colours balls, we get
\[ \Rightarrow {P_{WHITE}} = \dfrac{1}{3}\]

We will now find the probability of getting the black balls from the three colours balls, we get
\[ \Rightarrow {P_{BLACK}} = \dfrac{1}{3}\]
We know that 10 balls are selected.
Let us assume that the probability for selection induces at least one ball from each colour to be found is \[{\text{P}}\].
So, we have \[{\text{Red}} + {\text{Black}} + {\text{White}} = 10\], such that \[{\text{Red}} \geqslant 1\], \[{\text{White}} \geqslant 1\] and \[{\text{Black}} \geqslant 1\].
Now we will find the total number of ways for selecting from the above method, we get
\[ \Rightarrow {}^{7 + 3 - 1}{C_{3 - 1}} = {}^9{C_2}\]
We will find the total ways such that no constraints on quantity, we get
\[ \Rightarrow {}^{10 + 3 - 1}{C_{3 - 1}} = {}^{12}{C_2}\]
Dividing the total ways for selecting by the ways such that no constraints on quantity, we get
\[ \Rightarrow {\text{P}} = \dfrac{{{}^9{C_2}}}{{{}^{12}{C_2}}}\]
We will use the formula to calculate combinations in the above equation is \[{}^n{C_r} = \dfrac{{\left. {\underline {\,
 n \,}}\! \right| }}{{\left. {\underline {\,
 r \,}}\! \right| \cdot \left. {\underline {\,
 {n - r} \,}}\! \right| }}\], where \[n\] is the number of items, and \[r\] represents the number of items being chosen.
\[
   \Rightarrow {\text{P}} = \dfrac{{\dfrac{{9!}}{{2!\left( {9 - 2} \right)!}}}}{{\dfrac{{12!}}{{2!\left( {12 - 2} \right)!}}}} \\
   \Rightarrow {\text{P}} = \dfrac{{\dfrac{{9!}}{{2!7!}}}}{{\dfrac{{12!}}{{2!10!}}}} \\
   \Rightarrow {\text{P}} = \dfrac{{9 \times 8}}{{12 \times 11}} \\
   \Rightarrow {\text{P}} = \dfrac{6}{{11}} \\
 \]

Hence, option B is correct.

Note:
Student takes the total number of outcomes as 11 instead of 3, which is wrong. We will divide the obtained probabilities of the total ways for selecting by the ways such that no constraints on quantity, so not multiply them. Here, students must take care while simplifying the conditions given in the question into the combinations. Some students use the formula of permutation, \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] instead of combinations is \[{}^n{C_r} = \dfrac{{\left. {\underline {\,
 n \,}}\! \right| }}{{\left. {\underline {\,
 r \,}}\! \right| \cdot \left. {\underline {\,
 {n - r} \,}}\! \right| }}\], where \[n\] is the number of items, and \[r\] represents the number of items being chosen, which is wrong.