Question

The eighth-grade students at Keisel Middle School are divided as number of left-handed students and right-handed students by gender. Let there be $5$ times as many right-handed female students as there are left-handed female students, and there are $9$ times as many right-handed male students as there are left-handed male students. If there is a total of $18$ left-handed students and $122$ right-handed students in the school, which of the following is closest to the probability that a right-handed student selected at random is female? (Note: Assume that none of the eighth-grade students are both right-handed and left-handed.)

• A. $0.410$
• B. $0.357$
• C. $0.333$
• D. $0.250$

Answer 1

The correct option is A $0.410$

Let $x$ be the number of left handed female students and let $y$ be the number of left handed male students.

Then the number of right handed female students will be $5x$ and the number of right handed male students will be $9y$. Since the total number of left handed students is $18$ and the total number of right handed students is $122$, the following system of equations must be satisfied.

$x+y=18$ and $5x+9y=122$

Solving this system gives $x=10$ and $y=8$.

Thus, $50$ of the $122$ right handed students are female. Therefore, the probability that a right handed student selected at random is female is the fraction $\frac{50}{122}$, which to the nearest thousandth is $0.410$.