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

Hint: Here we use the concept of probability and find the probability in each case by keeping the number of given colored marbles as a favorable outcome and total number of marbles as total outcomes.
* Probability of an event is given by dividing the number of favorable outcomes by total number of outcomes.

Complete step-by-step answer:
We are given that there is a box that contains 5 red marbles, 8 white marbles and 4 green marbles.
Number of red marbles \[ = 5\]
Number of white marbles\[ = 8\]
Number of green marbles\[4\]
\[ \Rightarrow \]The total number of marbles in the box \[ = \]number of red marbles\[ + \]number of white marbles\[ + \]number of green marbles.
\[ \Rightarrow \]The total number of marbles in the box\[ = 5 + 8 + 4\]
\[ \Rightarrow \]The total number of marbles in the box\[ = 17\]..............… (1)
Now we solve for each part separately.
(i) Red
We have to find the probability that one marble taken out at random from the box is a red marble.
Number of red marbles \[ = 5\]
\[ \Rightarrow \]Number of favorable outcomes\[ = 5\]
The total number of marbles in the box\[ = 17\]
\[ \Rightarrow \]Number of total outcomes\[ = 17\]
Since, Probability\[ = \]Number of favorable outcomes \[/\]total number of outcomes
\[ \Rightarrow \]Probability\[ = \dfrac{5}{{17}}\]
\[\therefore \]Probability that the marble will be a red marble is \[\dfrac{5}{{17}}\] .
(ii) White
We have to find the probability that one marble taken out at random from the box is a white marble.
Number of white marbles \[ = 8\]
\[ \Rightarrow \]Number of favorable outcomes\[ = 8\]
The total number of marbles in the box\[ = 17\]
\[ \Rightarrow \]Number of total outcomes\[ = 17\]
Since, Probability\[ = \]Number of favorable outcomes \[/\]total number of outcomes
\[ \Rightarrow \]Probability\[ = \dfrac{8}{{17}}\]
\[\therefore \]Probability that the marble will be a white marble is \[\dfrac{8}{{17}}\].
(iii) Not green?
We have to find the probability that one marble taken out at random from the box is not a green marble.
Not a green marble means only red and white marbles.
Number of red marbles \[ = 5\]
Number of white marbles \[ = 8\]
Total number of marbles to choose from\[ = 5 + 8\]
Total number of marbles to choose from\[ = 13\]
\[ \Rightarrow \]Number of favorable outcomes\[ = 13\]
The total number of marbles in the box\[ = 17\]
\[ \Rightarrow \]Number of total outcomes\[ = 17\]
Since, Probability\[ = \]Number of favorable outcomes \[/\]total number of outcomes
\[ \Rightarrow \]Probability\[ = \dfrac{{13}}{{17}}\]
\[\therefore \]Probability that the marble will not be a green marble is \[\dfrac{{13}}{{17}}\].

Note: Alternative method:
We can solve this question using a combination method as well.
Formula of combination is given by\[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], where factorial is expanded by the formula \[n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1\]
If we substitute r as 1
\[{ \Rightarrow ^n}{C_1} = \dfrac{{n!}}{{(n - 1)!1!}}\]
Write numerator using factorial expansion
\[{ \Rightarrow ^n}{C_1} = \dfrac{{n(n - 1)!}}{{(n - 1)!1!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^n}{C_1} = n\].............… (2)
(i) Red
We have to find the probability that one marble taken out at random from the box is a red marble.
Number of red marbles \[ = 5\]
So, number of ways to choose 1 marble out of 5 marbles\[{ = ^5}{C_1}\]
The total number of marbles in the box\[ = 17\]
So, number of ways to choose 1 marble out of 17 marbles\[{ = ^{17}}{C_1}\]
Since, Probability\[ = \]Number of favorable outcomes \[/\]total number of outcomes
\[ \Rightarrow \]Probability\[ = \dfrac{{^5{C_1}}}{{^{17}{C_1}}}\]
Use equation (2)
\[ \Rightarrow \]Probability\[ = \dfrac{5}{{17}}\]
\[\therefore \]Probability that the marble will be a red marble is\[\dfrac{5}{{17}}\] .
(ii) White
We have to find the probability that one marble taken out at random from the box is a white marble.
Number of white marbles \[ = 8\]
So, number of ways to choose 1 marble out of 8 marbles\[{ = ^8}{C_1}\]
The total number of marbles in the box\[ = 17\]
So, number of ways to choose 1 marble out of 17 marbles\[{ = ^{17}}{C_1}\]
Since, Probability\[ = \]Number of favorable outcomes \[/\]total number of outcomes
\[ \Rightarrow \]Probability\[ = \dfrac{{^8{C_1}}}{{^{17}{C_1}}}\]
Use equation (2)
\[ \Rightarrow \]Probability\[ = \dfrac{8}{{17}}\]
\[\therefore \]Probability that the marble will be a white marble is\[\dfrac{8}{{17}}\] .
(iii) Not green?
We have to find the probability that one marble taken out at random from the box is not a green marble.
Not a green marble means only red and white marbles.
Number of red marbles \[ = 5\]
Number of white marbles \[ = 8\]
Total number of marbles to choose from\[ = 5 + 8\]
Total number of marbles to choose from\[ = 13\]
So, number of ways to choose 1 marble out of 13 marbles\[{ = ^{13}}{C_1}\]
The total number of marbles in the box\[ = 17\]
So, number of ways to choose 1 marble out of 17 marbles\[{ = ^{17}}{C_1}\]
Since, Probability\[ = \]Number of favorable outcomes \[/\]total number of outcomes
\[ \Rightarrow \]Probability\[ = \dfrac{{^{13}{C_1}}}{{^{17}{C_1}}}\]
Use equation (2)
\[ \Rightarrow \]Probability\[ = \dfrac{{13}}{{17}}\]
\[\therefore \]Probability that the marble will not be a green marble is \[\dfrac{{13}}{{17}}\] .