- Identify the total number of students as \(26\), with \(10\) boys and \(16\) non-boys (girls).
- Calculate the probability of selecting a girl first: \(P(G_1) = \frac{16}{26} = \frac{8}{13}\).
- Calculate the probability of selecting a second girl given the first: \(P(G_2|G_1) = \frac{15}{25} = \frac{3}{5}\).
- Multiply these probabilities to find the total probability: \(\frac{8}{13} \times \frac{3}{5} = \boxed{\frac{24}{65}}\).
Explanation
- Understand the problem and provided data
We start with a total of \(26\) students in the class. We are told that \(10\) of these students are boys. Since the remaining students are not boys, the number of girls (non-boys) is \(26 - 10 = 16\). Eduardo draws two names from the hat one by one without replacement. Our objective is to find the probability that both selected names are non-boys (girls).
- Calculate the sequential probabilities
We calculate the probability of each draw sequentially.
- The probability of selecting a girl on the first draw, \(P(G_1)\): There are \(16\) girls out of \(26\) total students, so \(P(G_1) = \frac{16}{26} = \frac{8}{13}\).
- The probability of selecting a girl on the second draw, given that the first was a girl, \(P(G_2 | G_1)\): After removing one girl, there are \(15\) girls left out of the remaining \(25\) total students, so \(P(G_2 | G_1) = \frac{15}{25} = \frac{3}{5}\).
- Calculate final probability and review options
To find the joint probability that both events occur (both are girls), we multiply the probability of the first event by the conditional probability of the second event:
\[P(G_1 \cap G_2) = P(G_1) \times P(G_2 | G_1) = \frac{8}{13} \times \frac{3}{5} = \frac{24}{65}\]
Comparing this result to the given options, we check for consistency. Note that the calculated probability is \(\frac{24}{65}\). Since this is not listed exactly as an option (a, b, c, or d), let's re-verify the input.
Wait, if the options were a) \(\frac{9}{65}\), b) \(\frac{24}{55}\), c) \(\frac{64}{169}\), d) \(\frac{128}{325}\)... wait, the calculation \(\frac{24}{65} = \frac{120}{325}\). Let me re-calculate or re-examine the choices.
Re-checking the multiplication: \(16/26 * 15/25 = (16*15)/(26*25) = 240/650 = 24/65\).
Is there any other interpretation? "10 of whom are boys". "26 other students". "Eduardo has to pull two names". Everything seems correct.
Perhaps there is a typo in the provided options, or I should double-check the fraction reduction. \(24/65\) is the correct mathematical result based on the provided numbers. Since no option matches \(\frac{24}{65}\), I will present the correct calculation clearly.
Examples
This type of probability calculation is essential for understanding selection processes without replacement, such as assigning partners in a classroom or picking winning raffle tickets from a limited pool. By understanding how the pool of available options changes after each selection, you can accurately predict the likelihood of specific outcomes in various real-world scenarios. For example, if you have a deck of cards and want to calculate the odds of drawing two consecutive hearts, you would use this same method: \(\frac{13}{52} \times \frac{12}{51} = \frac{156}{2652} = \frac{1}{17}\).