Question

There are 10 Mangoes and 6 Apples in a fruit basket in which 3 Mangoes and 2 Apples are not good. A person selected 2 fruits at random from the basket. Let P denotes probability that the person has drawn either apples or good fruits. Given that $yp - x = 0$, where, $x, y \in N$; HCF$(x, y) = 1$. Then which of the following is/are correct.

• A. 2x - y - 1 = 0
• B. 2x - y - 2 = 0
• C. |x - y| < 6
• D. |x - y| < 8

Answer 1

Answer: (C), (D)

|x - y| < 6, |x - y| < 8

Answer 2

Total fruits: 10 Mangoes + 6 Apples = 16

Not good Mangoes: 3, Good Mangoes: 7 Not good Apples: 2, Good Apples: 4 Total not good fruits: 5, Total good fruits: 11

Ways to select 2 fruits: $\binom{16}{2} = 120$

P = P(at least one Apple or at least one good fruit) $P = 1 - P(\text{no Apples and no good fruits}) = 1 - P(\text{both are not good Mangoes})$

Ways to select 2 not good Mangoes: $\binom{3}{2} = 3$

$P(\text{both are not good Mangoes}) = \frac{3}{120} = \frac{1}{40}$

$P = 1 - \frac{1}{40} = \frac{39}{40}$

Given $yp - x = 0$, so $p = \frac{x}{y} = \frac{39}{40}$. Thus, $x = 39$ and $y = 40$.

Option 1: $2x - y - 1 = 2(39) - 40 - 1 = 37 \neq 0$ (Incorrect)

Option 2: $2x - y - 2 = 2(39) - 40 - 2 = 36 \neq 0$ (Incorrect)

Option 3: $|x - y| = |39 - 40| = 1 < 6$ (Correct)

Option 4: $|x - y| = |39 - 40| = 1 < 8$ (Correct)