Question

A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red, (ii) white, (iii) not green?

Answer 1

In the solution, first we have to find the total numbers of marbles present in the box. After that we have to divide each of the marbles selected with the total numbers of marble for each case. This will give us the probability for selecting the marble in each case.

Complete step by step solution:
Given that the number of red marbles, white marbles and green marbles are 5, 8 and 4 respectively.
Thus the total number of marbles$\begin{array}{l} = 5 + 8 + 4\\\ = 17\end{array}$
Case (i) The probability that the marble taken out will be red.
Number of red marbles in the box is 5.
Let $P\left( {{\rm{red}}\;{\rm{marble}}} \right)$ the probability of the red marble taken out.
$\begin{array}{c}P\left( {{\rm{red}}\;{\rm{marble}}} \right) = \dfrac{{{\rm{number}}\;{\rm{of}}\;{\rm{red}}\;{\rm{marbles}}}}{{{\rm{total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{marbles}}}}\\\ = \dfrac{5}{{17}}\end{array}$

Case (ii) The probability that the marble taken out will be white.
Number of white marbles in the box is 8.
Let $P\left( {{\rm{white}}\;{\rm{marble}}} \right)$ the probability of the red marble taken out.
$\begin{array}{c}P\left( {{\rm{white}}\;{\rm{marble}}} \right) = \dfrac{{{\rm{number}}\;{\rm{of}}\;{\rm{white}}\;{\rm{marbles}}}}{{{\rm{total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{marbles}}}}\\\ = \dfrac{8}{{17}}\end{array}$

Case (iii) The probability that the marble taken out is not green.
Here, we have to calculate the probability of not selecting the green marble. For that we calculate the probability of selecting the green marbles. We have to subtract the probability of selecting the green ball from 1.

Number of green marbles in the box is 4.
Let $P\left( {{\rm{green}}\;{\rm{marble}}} \right)$ the probability of the green marble taken out.
$\begin{array}{c}P\left( {{\rm{green}}\;{\rm{marble}}} \right) = \dfrac{{{\rm{number}}\;{\rm{of}}\;{\rm{green}}\;{\rm{marbles}}}}{{{\rm{total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{marbles}}}}\\\ = \dfrac{4}{{17}}\end{array}$
Let $P\left( {{\rm{marble}}\;{\rm{taken}}\;{\rm{out}}\;{\rm{is}}\;{\rm{not}}\;{\rm{green}}} \right)$ the probability of the green marble for not taken out.
$\begin{array}{c}P\left( {{\rm{marble}}\;{\rm{taken}}\;{\rm{out}}\;{\rm{is}}\;{\rm{not}}\;{\rm{green}}} \right) = 1 - P\left( {{\rm{green}}\;{\rm{marble}}} \right)\\\ = 1 - \dfrac{4}{{17}}\\\ = \dfrac{{17 - 4}}{{17}}\\\ = \dfrac{{13}}{{17}}\end{array}$

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.