Question

A basket contains 20 apples and 10 oranges out of which 5 apples and 3 oranges are rotten. If a person takes out one fruit at random, find the probability that the fruit is either an apple or a good fruit.

Answer 1

Hint: Find the number of rotten and good fruits first. Now, consider two events that is, one is the event of selecting one apple and the other is the event of selecting one good fruit. Now, find the union of these two events by using $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$.

Complete step-by-step answer:
Here, we are given that a basket contains 20 apples and 10 oranges out of which 5 apples and 3 oranges are rotten. If a person takes out one fruit at random, we have to find the probability of the fruit being either an apple or a good fruit. So, we are given that there are 10 oranges, out of which 3 are rotten, so 7 are good fruits. Also, there are 20 apples, out of which 5 are rotten, so 16 are good fruits.
Fruits

Fruits	10 oranges	20 apples	Total
Rotten	3	5	8
Good fruits	7	15	22

Let us consider them as two events. The first event is selecting an apple and the second event as selecting a good fruit, so we get,
Event A = Event of selecting one apple. Event B = Event of selecting a good fruit.
Here we have to find the probability that the chosen fruit is either an apple or a good fruit, so we need to find $P\left( A\cup B \right)$.
We know that $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)\ldots \ldots \ldots \left( i \right)$
Now, let us find the probability of the event of selecting one apple $=P\left( A \right)$.
$P\left( A \right)=\dfrac{\text{Number of favourable event}}{\text{Number of total events}}=\dfrac{\text{Number of ways of choosing an apple}}{\text{Number of ways of choosing any fruit}}$.
We have 20 apples and the total number of fruits as 30, so, $P\left( A \right)=\dfrac{{}^{20}{{C}_{1}}}{{}^{30}{{C}_{1}}}$. Now we know the general formula is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . So, applying the formula, we get,
$\begin{align}
  & P\left( A \right)=\dfrac{\dfrac{20!}{1!\times 19!}}{\dfrac{30!}{1!\times 29!}} \\
 & P\left( A \right)=\dfrac{\dfrac{20\times 19!}{1!\times 19!}}{\dfrac{30\times 29!}{1!\times 29!}} \\
 & P\left( A \right)=\dfrac{\dfrac{20}{1}}{\dfrac{30}{1}} \\
 & P\left( A \right)=\dfrac{20}{30}\Rightarrow \dfrac{2}{3}\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
Now, let us find the probability of the event of selecting a good fruit $=P\left( B \right)$.
$P\left( B \right)=\dfrac{\text{Number of ways of choosing a good fruit}}{\text{Number of ways of choosing a fruit}}$. We have total of $7+15=22$ good fruits, so we get, $P\left( B \right)=\dfrac{{}^{22}{{C}_{1}}}{{}^{30}{{C}_{1}}}$. Now applying the formula again, we get,
$\begin{align}
  & P\left( B \right)=\dfrac{\dfrac{22!}{1!\times 21!}}{\dfrac{30!}{1!\times 29!}} \\
 & P\left( B \right)=\dfrac{\dfrac{22\times 21!}{1!\times 21!}}{\dfrac{30\times 29!}{1!\times 29!}} \\
 & P\left( B \right)=\dfrac{\dfrac{22}{1}}{\dfrac{30}{1}} \\
 & P\left( B \right)=\dfrac{22}{30}\Rightarrow \dfrac{11}{15}\ldots \ldots \ldots \left( iii \right) \\
\end{align}$
Now, $A\cap B$ is the event of selecting a good apple, so, $P\left( A\cap B \right)=\dfrac{\text{Number of ways of choosing a good apple}}{\text{Number of ways of choosing a fruit}}$. We have 15 good apples, so we have $P\left( A\cap B \right)=\dfrac{{}^{15}{{C}_{1}}}{{}^{30}{{C}_{1}}}$. Now again applying the formula, we get,
$\begin{align}
  & P\left( A\cap B \right)=\dfrac{\dfrac{15!}{1!\times 14!}}{\dfrac{30!}{1!\times 29!}} \\
 & P\left( A\cap B \right)=\dfrac{\dfrac{15\times 14!}{1!\times 14!}}{\dfrac{30\times 29!}{1!\times 29!}} \\
 & P\left( A\cap B \right)=\dfrac{\dfrac{15}{1}}{\dfrac{30}{1}} \\
 & P\left( A\cap B \right)=\dfrac{15}{30}\Rightarrow \dfrac{1}{2}\ldots \ldots \ldots \left( iv \right) \\
\end{align}$
Now substituting the values of $P\left( A \right),P\left( B \right),P\left( A\cap B \right)$ in equation (i), we get, $\begin{align}
  & P\left( A\cup B \right)=\dfrac{2}{3}+\dfrac{11}{15}-\dfrac{1}{2} \\
 & P\left( A\cup B \right)=\dfrac{20+22-15}{30} \\
 & P\left( A\cup B \right)=\dfrac{27}{30}\Rightarrow \dfrac{9}{10} \\
\end{align}$
Therefore, we get the probability that the fruit is either an apple or a good one as $\dfrac{9}{10}$.

Note: Here the students must note that in questions involving ‘either or’ we need to find the union of two events that is, $P\left( A\cup B \right)$ and for questions involving ‘and’ we need to find the intersection of the two events that is, $A\cap B$. The students must remember the formula for $P\left( A\cup B \right)$, which is, $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$ and find the unknown value by substituting the known values.