A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?
Solution
Correct Answer: Option C
\(n\left(s\right)={}^{30}C_2\)
Let A be the event of getting two oranges and
B be the event of getting two non-defective fruits.
and \(\left(A\cap B\right)\) be the event of getting two non-defective oranges
\(\therefore\;P\left(A\right)=\frac{{}^{20}C_2}{{}^{30}C_2},\;P\left(B\right)=\frac{{}^{22}C_2}{{}^{30}C_2}\;and\;P\left(A\cap B\right)=\frac{{}^{15}C_2}{{}^{30}C_2}\)
\(\therefore P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\)
= \(\frac{{}^{20}C_2}{{}^{30}C_2}+\frac{{}^{22}C_2}{{}^{30}C_2}-\frac{{}^{15}C_2}{{}^{30}C_2}=\frac{316}{435}\)
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