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A fruit basket contains 6 apples, x mangoes, x + 3 guavas. If a fruit is chosen randomly from the basket, then the probability of it being guava is 37 . What is the probability of choosing two fruits from the basket one after the other without replacement, such that the first fruit chosen is a mango and the second fruit chosen is an apple?

Question

A fruit basket contains 6 apples, x mangoes, x + 3 guavas. If a fruit is chosen randomly from the basket, then the probability of it being guava is 37 . What is the probability of choosing two fruits from the basket one after the other without replacement, such that the first fruit chosen is a mango and the second fruit chosen is an apple?

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Solution

First, let's find the value of x.

We know that the probability of choosing a guava is 3/7. The total number of fruits in the basket is 6 (apples) + x (mangoes) + x + 3 (guavas) = 2x + 9.

So, the probability of choosing a guava is (x + 3) / (2x + 9) = 3/7.

Solving this equation for x, we get x = 6.

So, the total number of fruits in the basket is 2*6 + 9 = 21.

Now, let's find the probability of choosing a mango first and then an apple.

The probability of choosing a mango first is 6/21 = 2/7.

After choosing a mango, the total number of fruits left in the basket is 20, and the number of apples is still 6.

So, the probability of choosing an apple next is 6/20 = 3/10.

Therefore, the probability of choosing a mango first and then an apple is (2/7) * (3/10) = 6/70 = 3/35.

This problem has been solved

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