Aptitude Discussion

Q. |
There are 2 positive integers $x$ and $y$. What is the probability that $x+y$ is odd? |

✖ A. |
1/4 |

✖ B. |
1/3 |

✔ C. |
1/2 |

✖ D. |
1/5 |

**Solution:**

Option(**C**) is correct

Here we can have four cases:

**a).** $x$ is even, $y$ is even

**b).** $x$ is odd, $y$ is even

**c).** $x$ is even, $y$ is odd

**d).** $x$ is odd, $y$ is odd

Out of these four cases, in case '$b$' and '$c$' the sum will be odd.

So required probability = 2/4 = **1/2**

**Edit:** For further analysis on this question or to deduce generalised results check comment by **Saurabh Bansal.**

**Saurabh Bansal**

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**Sohameddy**

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Guess, there is an error. All along, I was thinking of 2 digit numbers that are made up of x and y.

**Sohameddy**

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I think all the choices given are wrong. x and y are positive integers. So, x and y can take values from 1,2,3 till 9.

To get odd sum, we need to consider two cases of first digit even, second digit odd or vice versa.

Bear in mind that there are 5 odd positive integers and 4 even positive integers.

Hence, no. of possible Even Odd numbers = 4x 5 = 20

And no.of possible Odd Even numbers = 5 X 4 = 20

Total = 40

Altogether there are 9 X 9 = 81 numbers from 11 to 99 minus the two digits numbers that end with 0 ( zero)

Thus, the probability is 40/81.

I guess, you are thinking something else. Question talks about the SUM of two positive integers and not making a new number by combining them.

To me 1/2 is the correct answer and solution given is justified.

Simpler, way to think is that sum of two positive numbers is a positive number. A positive number has an equal probability of being odd or even, hence the answer is 1/2.

Same is true for many mathematical operations like $x+y$, $x \times y$, $x-y$ as long as $x$ and $y$ are used symmetrically in the equation.

So $x+2\times y$ might have a different answer than 1/2.