Question

A bag contains 6 R, 4 W and 8 B balls. If 3 balls are drawn simultaneously at random determine the probability of the event
(1) All 3 are red
(2) All 3 are black
(3) At least 1 is red
(4) 1 of each colour are drawn
(5) The balls are drawn in the order of red, white, and black.
(6) 2 are white and 1 is red.

Answer 1

In the solution, first we have to find the total numbers of balls present in the bag. After that we have to divide each of the balls selected with the total numbers of balls contained in the bag, it will give the probability for selecting the balls in each event.

Complete step by step solution:
Total number of balls in the bag
$\begin{array}{c} = 6 + 4 + 8\\\ = 18\end{array}$
It is given that the total number of balls drawn from the bag is 3.

Now we have to find the probability for selecting 3 balls from 18 balls
$\begin{array}{l}P\left( {{\rm{selecting}}\;{\rm{three}}\;{\rm{balls}}} \right) = {}^{18}{C_3}\\\ \Rightarrow P\left( {{\rm{selecting}}\;{\rm{three}}\;{\rm{balls}}} \right) = \dfrac{{18 \cdot 17 \cdot 16}}{{3 \cdot 2 \cdot 1}}\\\ \Rightarrow P\left( {{\rm{selecting}}\;{\rm{three}}\;{\rm{balls}}} \right) = \dfrac{{4696}}{6}\\\ \Rightarrow P\left( {{\rm{selecting}}\;{\rm{three}}\;{\rm{balls}}} \right) = 816\end{array}$

Case (1) when all 3 balls are red.
Given, the number of red balls are 6
Now probability of the red ball drawn.
$\begin{array}{l}P\left( {{\rm{red}}\;{\rm{ball}}} \right) = {}^6{C_3}\\\ \Rightarrow P\left( {{\rm{red}}\;{\rm{ball}}} \right) = \dfrac{{6 \cdot 5 \cdot 4}}{{3 \cdot 2 \cdot 1}}\\\ \Rightarrow P\left( {{\rm{red}}\;{\rm{ball}}} \right) = \dfrac{{120}}{6}\\\ \Rightarrow P\left( {{\rm{red}}\;{\rm{ball}}} \right) = 20\end{array}$

Now, we can find the probability that all 3 balls are red
$\begin{array}{l}P\left( {{\rm{all}}\;{\rm{red}}} \right) = \dfrac{{P\left( {{\rm{red}}\;{\rm{ball}}} \right)}}{{P\left( {{\rm{selecting}}\;{\rm{three}}\;{\rm{balls}}} \right)}}\\\ \Rightarrow P\left( {{\rm{all}}\;{\rm{red}}} \right) = \dfrac{{20}}{{816}}\\\ \Rightarrow P\left( {{\rm{all}}\;{\rm{red}}} \right) = \dfrac{5}{{204}}\end{array}$

Case (2) when all 3 balls are black
Given, the number of black balls are 8
Now probability of the red ball drawn.
$\begin{array}{l}P\left( {{\rm{black}}\;{\rm{ball}}} \right) = {}^8{C_3}\\\ \Rightarrow P\left( {{\rm{black}}\;{\rm{ball}}} \right) = \dfrac{{8 \cdot 7 \cdot 6}}{{3 \cdot 2 \cdot 1}}\\\ \Rightarrow P\left( {{\rm{black}}\;{\rm{ball}}} \right) = \dfrac{{336}}{6}\\\ \Rightarrow P\left( {{\rm{black}}\;{\rm{ball}}} \right) = 56\end{array}$

Now, we can find the probability that all 3 balls are black
$\begin{array}{l}P\left( {{\rm{all}}\;{\rm{black}}} \right) = \dfrac{{P\left( {{\rm{black}}\;{\rm{ball}}} \right)}}{{P\left( {{\rm{selecting}}\;{\rm{three}}\;{\rm{balls}}} \right)}}\\\ \Rightarrow P\left( {{\rm{all}}\;{\rm{black}}} \right) = \dfrac{{56}}{{816}}\\\ \Rightarrow P\left( {{\rm{all}}\;{\rm{black}}} \right) = \dfrac{7}{{102}}\end{array}$

Case (3) probability for at least 1 is red
Probability of getting at least 1 red ball = Probability of getting 1 red ball + Probability of getting 2 red balls + Probability of getting 3 red balls
Probability of getting 1 red ball $= \dfrac{{{}^6{C_1} \cdot {}^{12}{C_2}}}{{{}^{18}{C_3}}}$
$\begin{array}{l} = \dfrac{{6 \times 66}}{{\dfrac{{18 \cdot 17 \cdot 16}}{{3 \cdot 2 \cdot 1}}}}\\\ = \dfrac{{396}}{{816}}\end{array}$

Probability of getting 2 red ball $= \dfrac{{{}^6{C_2} \cdot {}^{12}{C_1}}}{{{}^{18}{C_3}}}$
$\begin{array}{l} = \dfrac{{15 \times 12}}{{\dfrac{{18 \cdot 17 \cdot 16}}{{3 \cdot 2 \cdot 1}}}}\\\ = \dfrac{{15}}{{68}}\end{array}$

Probability of getting 3 red ball $= \dfrac{{{}^6{C_3} \cdot {}^{12}{C_0}}}{{{}^{18}{C_3}}}$
$= \dfrac{5}{{204}}$

By adding all three probability, we get the probability of getting at least one red ball.
Probability of getting at least 1 red ball
$\begin{array}{l} = \dfrac{{396}}{{816}} + \dfrac{{15}}{{68}} + \dfrac{5}{{204}}\\\ = \dfrac{{596}}{{816}}\\\ = \dfrac{{149}}{{204}}\end{array}$

Thus, the probability of getting at least 1 red ball is$\dfrac{{149}}{{204}}$.

Case (4) probability for 1 ball of each colour are drawn
Let probability of getting one ball from each colour is P(1 ball from each colour)
$\therefore$ P(1 ball from each colour)
$\begin{array}{c} = \dfrac{{{}^4{C_1} \cdot {}^6{C_1} \cdot {}^8{C_1}}}{{{}^{18}{C_3}}}\\\ = \dfrac{{4 \times 6 \times 8}}{{\dfrac{{18 \cdot 17 \cdot 16}}{{3 \cdot 2 \cdot 1}}}}\\\ = \dfrac{4}{{17}}\end{array}$
Thus, the probability of getting one ball from each colour is$= \dfrac{4}{{17}}$.
Case (5) when the balls are drawn in the order of red, white, black.
It is given that the balls are drawn in the order of red, white, black.
Therefore, the probability for the red ball
$\begin{array}{l} = \dfrac{{{}^4{C_1}}}{{{}^{18}{C_1}}}\\\ = \dfrac{4}{{18}}\end{array}$
After drawing the red ball, 17 balls were left in the bag.
Therefore, the probability for the white ball
$\begin{array}{l} = \dfrac{{{}^6{C_1}}}{{{}^{17}{C_1}}}\\\ = \dfrac{6}{{17}}\end{array}$
After drawing the red ball and white ball, 16 balls were left in the bag.
Therefore, the probability for the black ball
$\begin{array}{l} = \dfrac{{{}^8{C_1}}}{{{}^{16}{C_1}}}\\\ = \dfrac{8}{{16}}\end{array}$

Thus, the probability that the balls are drawn in the order of red, white, black $\begin{array}{l} = \dfrac{4}{{18}} \times \dfrac{6}{{17}} \times \dfrac{8}{{16}}\\\ = \dfrac{2}{{51}}\end{array}$

Case (6) probability when 2 balls are white and 1 ball is red.
Probability of getting two white ball and one red ball
$\begin{array}{l} = \dfrac{{{}^4{C_2} \cdot {}^6{C_1}}}{{{}^{18}{C_3}}}\\\ = \dfrac{{6 \times 6}}{{\dfrac{{18 \cdot 17 \cdot 16}}{{3 \cdot 2 \cdot 1}}}}\\\ = \dfrac{3}{{68}}\end{array}$

Hence, the probability of getting two white balls and one red ball is $\dfrac{3}{{68}}$.

Note: Probability explains the chances of happening in an even. Here we have to determine the probability for each event. Since the given cases are a combination. Thus, calculating the probability of the combination of each event, we get our required answers.