Question

A Group Housing Society has 600 members who have their houses on the campus and decided to hold a tree plantation drive on the occasion of New Year. Each household was given the choice of planting and sapling of its choice. The numbers of different types of sapling planted were \[\]
(i) Neem -125\[\]
(ii)Peepal-165\[\]
(iii) Creepers- 50\[\]
(iv)Fruit plant -150\[\]
(v) Flowering plant-110\[\]
On the opening ceremony one of the plants is selected randomly for a prize, after reading the above passage answer the following question what is the probability that the selected plant is\[\]
(a)A Fruit plant or flowering plant\[\]
(b) Either a Neem plant or Peepal plant\[\]

Answer 1

Hint: We find the number of saplings of each plant and find the total number of outcomes $n\left( S \right)$. We add the number of saplings of fruit plant and flowering plant to get the number of favourable outcomes $n\left( A \right)$ for part (a). We add the number of saplings of Neem and Peepal plant to get the number of favourable outcomes $n\left( A \right)$ for part (b). We can find the required probabilities $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$.\[\]

Complete step by step answer:
We know from definition of probability that if there is $n\left( A \right)$ number of ways of event $A$ occurring (or number of favourable outcomes) and $n\left( S \right)$ is the size of the sample space (number of all possible outcomes) then the probability of the event $A$ occurring is given by
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}\]
We are given the question that the number of sampling plants of Neem is 125, of Peepal is 165, of creeper is 50, of fruit plant is 150 and of flowering plant is 110. So the total number of plants is
\[125+\text{ }165\text{ }+\text{ }50\text{ }+\text{ }150\text{ }+\text{ }110\text{ }=\text{ }600\]
Each plant was given to a single household and the household planted in the New Year tree plantation Drive. Since there are 600 households and 600 plants then the total number of outcomes is
\[n\left( S \right)=600\]
We are further given in the question that On the opening ceremony one of the plants is selected randomly for a prize.
(a) We are asked to find the probability that the selected plant is a fruit plant or flowering plant. Let us denote the event of getting a fruit plant or flowering plant is $A$. So the total number of favourable outcomes is the sum of the number of saplings of the fruit plant and flowering plant that is $n\left( A \right)=150+110=260$. Then the required probability is
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{260}{600}=\dfrac{13}{20}\]
(b) (a) We are asked to find the probability that the selected plant is either a Neem plant or a Peepal plant. Let us denote the event of getting a Neem plant or a Peepal plant as$A$. So the total number of favourable outcomes is the sum of the number of saplings of a Neem plant or a Peepal plant that is $n\left( A \right)=165+125=290$.

Then the required probability is
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{290}{600}=\dfrac{29}{20}\].

Note: We note that the question presumes the saplings are distinct and the plants are distinct, for example a flowering plant cannot be a fruit plant or a creeper. If they would not have distinct we have to find number of saplings that are common, for example if there are $n\left( X \right)$ number of fruit plants and $n\left( Y \right)$ number of flowering plants then number of of flowering and fruit plants is given by $n\left( X\bigcap Y \right)$. Then the number of favourable outcomes $n\left( A \right)=n\left( X \right)+n\left( Y \right)-n\left( X\bigcap Y \right)$.