Question

Robert’s mother lets him pick one candy from a bag. He can’t see the candies. The number of candies of each color in the bag is shown in the following graph.

What is the probability that Robert will pick a red candy?

Answer 1

First find the total number of candies in a bag by adding the numbers of candies of each kind. After that use, the formula of probability $P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$ to find out the probability that the candy picked is red.

Formula Used:
Probability of an event is given by dividing the number of favorable outcomes by the total number of outcomes.
$P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$

Complete step by step answer:
Given: - A bag of candy of different colors.
The number of red candies in the bag is 6.
The number of orange candies in the bag is 5.
The number of yellow candies in the bag is 3.
The number of green candies in the bag is 3.
The number of blue candies in the bag is 2.
The number of pink candies in the bag is 4.
The number of purple candies in the bag is 2.
The number of brown candies in the bag is 5.
The total number of candies in the bag is given by the addition of the number of candies of each type.
$\Rightarrow$Total number of candies in a bag $=6+5+3+3+2+4+2+5$
$\Rightarrow$Total number of candies in a bag $=30$
Now we have to find the probability of one candy that is picked at random to be red.
The number of red candies in the bag is 6
So, the number of favorable outcomes is 6
The total number of outcomes is 30
Substitute the values in the formula for probability i.e. number of favorable outcomes divided by the total number of outcomes.
$\Rightarrow$Probability $=\dfrac{6}{30}$
Cancel out the common factors,
$\Rightarrow$Probability $=\dfrac{1}{5}$

Hence, the probability that Robert will pick a red candy is $\dfrac{1}{5}$.

Note:
Alternative method:
We can find the probability using a combination method as well i.e. $^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}$where n is the total number of choices available and r is the number of choices we have to make.
The number of red candies in the bag is 6.
The number of orange candies in the bag is 5.
The number of yellow candies in the bag is 3.
The number of green candies in the bag is 3.
The number of blue candies in the bag is 2.
The number of pink candies in the bag is 4.
The number of purple candies in the bag is 2.
The number of brown candies in the bag is 5.
The total number of candies in the bag is given by the addition of the number of players of each type.
$\Rightarrow$Total number of candies in a bag $=6+5+3+3+2+4+2+5$
$\Rightarrow$Total number of candies in a bag $=30$
Number of ways to choose 1 red candy from 6 red candy ${{=}^{6}}{{C}_{1}}$
Number ways to choose 1 candy from 30 candies ${{=}^{30}}{{C}_{1}}$
The probability that the chosen candy is red is given by dividing the number of ways to choose one all-rounder from all-rounders divided by the number of ways to choose one all-rounder from the total number of players.
$\Rightarrow$Probability$=\dfrac{^{6}{{C}_{1}}}{^{30}{{C}_{1}}}$ … (1)
Use the formula for combination to open the value in RHS$^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}$
Then $^{6}{{C}_{1}}=\dfrac{6!}{(6-1)!1!}$
Open the factorial in numerator using the formula $n!=n(n-1)!$
${{\Rightarrow }^{6}}{{C}_{1}}=\dfrac{6\times 5!}{5!1!}$
Cancel the same terms from numerator and denominator and substitute the value of $1!=1$in the denominator.
${{\Rightarrow }^{6}}{{C}_{1}}=6$
Similarly,$^{30}{{C}_{1}}=\dfrac{30!}{(30-1)!1!}$
Open the factorial in numerator using the formula $n!=n(n-1)!$
${{\Rightarrow }^{30}}{{C}_{1}}=\dfrac{30\times 29!}{29!1!}$
Cancel the same terms from numerator and denominator and substitute the value of $1!=1$in the denominator.
${{\Rightarrow }^{30}}{{C}_{1}}=30$
Substitute the values in equation (1)
$\Rightarrow$Probability$=\dfrac{6}{30}$
Cancel out the common factors,
$\Rightarrow$Probability $=\dfrac{1}{5}$
Hence, the probability that Robert will pick a red candy is $\dfrac{1}{5}$.