Question 10 Exercise 6.5

Solutions of Question 10 of Exercise 6.5 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

A basket contains $20$ apples and $10$ oranges out of which $5$ apples and $3$ oranges are defective.

If a person takes out $2$ at random what is the probability that either both are apples or both are good?

Total number of Apples $=20$ number of Oranges $=10$

number of defective apples $=5$

number of defective oranges $=3$.

Totál good apples $=15$

Defective apples $=5$

Total good oranges $=10$

number of defective oranges $=3$

number of good fruits $=22$

Now two fruits are chosen at random and we have to find the probability that either both are apples or both are non defective:

Let $E$ be the event that either both are apples or both are non defective.

$a A$ be the event that both are apples.

$B$ be the event that both are non defective.

Total number of ways taking $2$ fruits at random are: \begin{align}n(S)&={ }^{30} C_2\\ &=435\\ P(A)&=\dfrac{^{20} C_2}{^{30} C_2}\\ &=\dfrac{190}{435}=\dfrac{38}{87}\\ P(B)&=\dfrac{^{22} C_2}{^{30} C_2}\\ &=\dfrac{231}{435}=\dfrac{77}{145}\end{align} Also probability that both are nondefective apples is: \begin{align}P(\cap B)&=\dfrac{^{13} C_2}{^{30} C_2}\\ &=\dfrac{105}{435}=\dfrac{7}{29}\end{align} Now by addition law of probability, the probability that either both are apples or both are good is: \begin{align} P(A \cup B)&=P(A)+P(B)-P(A \cap B) \\ \Rightarrow P(A \cup B)&=\dfrac{190}{435}+\dfrac{231}{435}-\dfrac{105}{435} \\ \Rightarrow P(A \cup B)&=\dfrac{190+231-105}{435} \\ \Rightarrow P(A \cup B)&=\dfrac{316}{435}\end{align}

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