Three Bananas and three oranges are kept in a box. If two fruits are chosen at random, Find the probability that one is Banana and another one is orange?a)1/5b)3/5c)4/5d)2/5e)None of theseCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Problem: Three Bananas and three oranges are kept in a box. If two fruits are chosen at random, Find the probability that one is Banana and another one is orange?

Solution:

Total number of ways of choosing 2 fruits out of 6 = 6C2 = (6*5)/(2*1) = 15

Let A be the event of choosing one Banana and one Orange.

Number of ways of choosing one Banana out of three = 3C1 = 3

Number of ways of choosing one Orange out of three = 3C1 = 3

Therefore, number of ways of choosing one Banana and one Orange = 3*3 = 9

P(A) = Number of ways of choosing one Banana and one Orange / Total number of ways of choosing 2 fruits out of 6

P(A) = 9/15 = 3/5

Therefore, the probability that one is Banana and another one is orange is 3/5.

Answer: Option (b) 3/5.

Answer

Total probability = 6C2 = 15
Probability that one is Banana and another one is orange = 3C1 * 3C1 = 9
probability = 9/15 = 3/5

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