Question

A fruit basket contains 10 guavas and 20 bananas out of which 3 guavas and 5 bananas are defective. If two fruits selected at random, what is the probability that either both are bananas or both are non-defective?

Answer 1

n(s)=C230

 

 Let A be the event of getting two oranges and 

 

 B be the event of getting two non-defective fruits.

 

 and (A∩B) be the event of getting two non-defective oranges

 

 ∴ P(A)=C220C230, P(B)=C222C230 and P(A∩B)=C215C230

 

 ∴P(A∪B)=P(A)+P(B)-P(A∩B)

 

= C220C230+C222C230-C215C230=316435

Answer 2

Answer: Our required answer would be 0.967.

Step-by-step explanation:

Since we have given that

Number of guavas = 10

Number of bananas = 20

Number of defective guavas = 3

Number of defective bananas = 5

Total defectives = 3+5=8

Total non defectives = 30-8 =22

We need to find the probability that either both are bananas or both are non defective.

P(both bananas) + P(both non defective)

$\dfrac{^{20}C_2}{^{30}C_2}+\dfrac{^{22}C_2}{^{30}C_2}=0.436+0.531=0.967$

Hence, our required answer would be 0.967.