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Question: Sal has a small bag of candy containing three green candies and two red candies. While waiting for t...

Sal has a small bag of candy containing three green candies and two red candies. While waiting for the bus, he ate two candies out of the bag, one after another, without looking. What is the probability that both candies were the same color?

Explanation

Solution

Hint:Here, we will use the concept of probability as well as the concept of combinations. Probability of any event is given as P=number of favourable outcomes of the eventtotal number of events \text{P=}\dfrac{\text{number of favourable outcomes of the event}}{\text{total number of events }}. The number of ways of choosing K objects from a total of n objects at a time is given as nCk=n!k!×(n−k)!^{n}{{C}_{k}}=\dfrac{n!}{k!\times \left( n-k \right)!}.

Complete step-by-step answer:
Probability is concerned with numerical description of how likely an event is to occur or how likely it is that a proposition is true. The probability of an event is always a number between 0 and 1, where roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty. The higher the probability of an event, the more likely it is that the event will occur.
The total number of ways of selecting or K objects from a set of objects is given as:
nCk=n!k!×(n−k)!.......(1)^{n}{{C}_{k}}=\dfrac{n!}{k!\times \left( n-k \right)!}.......\left( 1 \right)
Here, we have been given that the set has three green candies and two red candies.
So, the total number of candies he has = 3+2 = 5.
Total number of ways in which he can choose 2 candies from these candies will be given as per equation (1) as:
5C2=5!2!×(5−2)! ⇒5C2=5!2!×3! ⇒5C2=5×4×3!2×3!=5×42=10 \begin{aligned} & ^{5}{{C}_{2}}=\dfrac{5!}{2!\times \left( 5-2 \right)!} \\\ & {{\Rightarrow }^{5}}{{C}_{2}}=\dfrac{5!}{2!\times 3!} \\\ & {{\Rightarrow }^{5}}{{C}_{2}}=\dfrac{5\times 4\times 3!}{2\times 3!}=\dfrac{5\times 4}{2}=10 \\\ \end{aligned}
Since it is given that the two candies that he ate were of the same color. This means that either he ate 2 green candies or 2 red candies.
Total number of ways of choosing 2 red candies from a total of 2 red candies =2C2=1^{2}{{C}_{2}}=1
Similarly number of ways of choosing 2 green candies from a total of 3 green candies = 3C2=3!2!×(3−2)!=3^{3}{{C}_{2}}=\dfrac{3!}{2!\times \left( 3-2 \right)!}=3
Now, the probability that both the candies were of same color is given as:
=Probability that both candies were red + probability that both candies were green
=number of ways of choosing 2 red candies total number of ways of choosing 2 candies +number of ways of choosing 2 green candiestotal number of ways of choosing 2 candies\dfrac{\text{number of ways of choosing 2 red candies }}{\text{total number of ways of choosing 2 candies }}+\dfrac{\text{number of ways of choosing 2 green candies}}{\text{total number of ways of choosing 2 candies}}
= 110+310=410=25\dfrac{1}{10}+\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}
Hence, the probability that both the candies that Sal ate were of the same color is = 25\dfrac{2}{5}.

Note: Students should note here that the value of 0! is equal to 1. That is why the value of 2C2^{2}{{C}_{2}} comes out to be 1.Students should remember combination formula nCk=n!k!×(n−k)!^{n}{{C}_{k}}=\dfrac{n!}{k!\times \left( n-k \right)!} for choosing k objects from n objects.Here in this question we have two cases for selection of candies to be the same which are totally independent events so we should add both probabilities to get the required answer.