Question

Suppose coloured balls are distributed in three boxes are as follows:

Colour $ \downarrow $	Box $ 1 $	Box $ 2 $	Box $ 3 $
RedWhiteBlue	$ \begin{gathered} 2 \\ 3 \\ 5 \\ \end{gathered} $	$ \begin{gathered} 4 \\ 1 \\ 3 \\ \end{gathered} $	$ \begin{gathered} 3 \\ 4 \\ 3 \\ \end{gathered} $
Total	$ 10 $	$ 8 $	$ 10 $

A box is selected at random from which a ball is selected at random. What is the probability that the ball is selected of red colour?

Answer 1

Hint: Here we will find the probability of the red balls from all the three boxes and then will find the total of all the three probabilities by taking addition to get the probability that the ball selected is of red colour irrespective of the selected box.

Complete step by step solution:
Probability is defined as the ratio of the favorable outcomes with the total number of possible outcomes.
Here, we have three boxes so when finding the probability for the red colour from any one box will be the factor of $ \dfrac{1}{3} $
Now, the probability of the red ball from the box one is given by $ = \dfrac{1}{3} \times \dfrac{2}{{10}} $
  ( $ \dfrac{2}{{10}} $ is taken for the favorable red balls upon the total number of balls in the box one)
Simplify the above expression finding the product of the terms –
the probability of the red ball from the box one $ = \dfrac{2}{{30}} $ …. (A)
Similarly,
The probability of the red ball from the box two $ = \dfrac{1}{3} \times \dfrac{4}{8} $
Simplify the above expression finding the product of the terms –
The probability of the red ball from the box two $ = \dfrac{4}{{24}} $ …. (B)
The probability of the red ball from the box three $ = \dfrac{1}{3} \times \dfrac{3}{{10}} $
Simplify the above expression finding the product of the terms –
The probability of the red ball from the box three $ = \dfrac{3}{{30}} $ …. (C)
The probability that the selected ball is of red colour $ = \dfrac{2}{{30}} + \dfrac{4}{{24}} + \dfrac{3}{{30}} $
Take LCM (least common multiple) for the above expression –
 $ = \dfrac{{2(4) + 4(5) + 3(4)}}{{120}} $
Simplify the above expression –
 $ = \dfrac{{8 + 20 + 12}}{{120}} $
 $ = \dfrac{{40}}{{120}} $
Common factors from the numerator and the denominator cancels each
 $ = \dfrac{1}{3} $
Hence, the probability that the ball selected of red color is $ \dfrac{1}{3} $
So, the correct answer is “ $ \dfrac{1}{3} $ ”.

Note: Be careful while solving and consider all the boxes for the probability. Always double check all the possible outcomes before placing them in the formula. Always remember that the range of the final answer of the probability is between zero and one. The probability can never be negative.