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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?
- a)315/435
- b)313/435
- c)317/435
- d)316/435
- e)None of these
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A fruit basket contains 10 Guavas and 20 Bananas out of which 3 Guavas...
P(A) = 20c2 / 30c2, P(B) = 22c2 / 30c2
P(A∩B) = 15c2 / 30c2
P(A∪B) = P(A) + P(B) – P(A∩B) ⇒ (20c2/30c2)+(22c2/30c2)- (15c2/30c2)=316/435
P(A∩B) = 15c2 / 30c2
P(A∪B) = P(A) + P(B) – P(A∩B) ⇒ (20c2/30c2)+(22c2/30c2)- (15c2/30c2)=316/435
Most Upvoted Answer
A fruit basket contains 10 Guavas and 20 Bananas out of which 3 Guavas...
Given:
- A fruit basket containing 10 Guavas and 20 Bananas.
- Out of which 3 Guavas and 5 Bananas are defective.
To find:
- Probability of getting either both Bananas or both non-defective fruits when two fruits are selected randomly.
Solution:
Total number of fruits = 10 + 20 = 30
Total number of defective fruits = 3 + 5 = 8
Total number of non-defective fruits = 30 - 8 = 22
Probability of selecting two Bananas = (20/30) * (19/29) = 38/87
Probability of selecting two non-defective fruits = (22/30) * (21/29) = 77/145
Now, we need to find the probability of either both Bananas or both non-defective fruits. This can be done using the formula:
P(A or B) = P(A) + P(B) - P(A and B)
where,
P(A) = probability of selecting two Bananas
P(B) = probability of selecting two non-defective fruits
P(A and B) = probability of selecting two Bananas that are also non-defective
To find P(A and B), we can use the formula:
P(A and B) = P(A) * P(B|A)
where,
P(B|A) = probability of selecting a non-defective Banana given that the first fruit selected was a Banana
= (15/29)
Therefore,
P(A and B) = (20/30) * (15/29) = 100/174
Now, substituting the values in the formula for P(A or B), we get:
P(A or B) = (38/87) + (77/145) - (100/174)
= 316/435
Therefore, the correct answer is option (D) 316/435.
- A fruit basket containing 10 Guavas and 20 Bananas.
- Out of which 3 Guavas and 5 Bananas are defective.
To find:
- Probability of getting either both Bananas or both non-defective fruits when two fruits are selected randomly.
Solution:
Total number of fruits = 10 + 20 = 30
Total number of defective fruits = 3 + 5 = 8
Total number of non-defective fruits = 30 - 8 = 22
Probability of selecting two Bananas = (20/30) * (19/29) = 38/87
Probability of selecting two non-defective fruits = (22/30) * (21/29) = 77/145
Now, we need to find the probability of either both Bananas or both non-defective fruits. This can be done using the formula:
P(A or B) = P(A) + P(B) - P(A and B)
where,
P(A) = probability of selecting two Bananas
P(B) = probability of selecting two non-defective fruits
P(A and B) = probability of selecting two Bananas that are also non-defective
To find P(A and B), we can use the formula:
P(A and B) = P(A) * P(B|A)
where,
P(B|A) = probability of selecting a non-defective Banana given that the first fruit selected was a Banana
= (15/29)
Therefore,
P(A and B) = (20/30) * (15/29) = 100/174
Now, substituting the values in the formula for P(A or B), we get:
P(A or B) = (38/87) + (77/145) - (100/174)
= 316/435
Therefore, the correct answer is option (D) 316/435.
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Community Answer
A fruit basket contains 10 Guavas and 20 Bananas out of which 3 Guavas...
P(A) = 20c2 / 30c2, P(B) = 22c2 / 30c2
P(A∩B) = 15c2 / 30c2
P(A∪B) = P(A) + P(B) – P(A∩B) ⇒ (20c2/30c2)+(22c2/30c2)- (15c2/30c2)=316/435
P(A∩B) = 15c2 / 30c2
P(A∪B) = P(A) + P(B) – P(A∩B) ⇒ (20c2/30c2)+(22c2/30c2)- (15c2/30c2)=316/435
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