Question

Out of the given 25 consecutive positive integers, three integers are drawn. If the least integer among given 25 integers is an odd number, then the probability that the sum of the three integers drawn is an even number is

• A. $289/575$
• B. $286/575$
• C. $288/575$
• D. $287/575$

Hint:

Remember the parity rules for sums: E+E=E, O+O=E, E+O=O. For a sum of three numbers to be even, the possibilities are (E,E,E) or (O,O,E).

Answer 1

The Correct Option is A

Since there are 25 consecutive integers and the least integer is odd, the sequence will be Odd, Even, Odd, Even, ...
In 25 integers, there will be 13 odd integers (O) and 12 even integers (E).
We are drawing 3 integers from these 25. The total number of ways to do this is $\binom{25}{3}$.
Total outcomes = $\binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 25 \times 4 \times 23 = 2300$.
The sum of the three integers drawn must be an even number. This can happen in two mutually exclusive cases:
Case 1: All three integers are even (E + E + E = Even).
The number of ways to choose 3 even integers from the 12 available is $\binom{12}{3}$.
$\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220$.
Case 2: One integer is even and two integers are odd (E + O + O = Even).
The number of ways to choose 1 even from 12 and 2 odd from 13 is $\binom{12}{1} \times \binom{13}{2}$.
$\binom{12}{1} \times \binom{13}{2} = 12 \times \frac{13 \times 12}{2} = 12 \times 13 \times 6 = 936$.
Total number of favorable outcomes = (Ways for Case 1) + (Ways for Case 2) = $220 + 936 = 1156$.
The probability is the ratio of favorable outcomes to total outcomes.
Probability = $\frac{1156}{2300}$.
Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4.
Probability = $\frac{1156 \div 4}{2300 \div 4} = \frac{289}{575}$.