Question

1. chloe was given the following problem in her statistics class: \a pencil case has seven pencils and three pens in it. suppose you pick two writing utensils together at random. what is the probability that both are pencils?\ chloe drew a 10x10 grid to represent the sample space of all combinations of writing utensils that could be chosen. she then counted 49 outcomes that included selecting two pencils. she concluded that the probability of picking two pencils is \\( \frac{49}{100} \\). explain why chloe is incorrect.

Explanation:

Chloe's grid assumes independence (like picking with replacement), but when picking two writing utensils together (without replacement), the sample - space is not \(10\times10\). The correct number of ways to choose 2 items out of \(n = 7 + 3=10\) is given by the combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\), where \(n = 10\) and \(k = 2\). The number of ways to choose 2 pencils out of 7 is \(C(7,2)=\frac{7!}{2!(7 - 2)!}=\frac{7\times6}{2\times1}=21\), and the number of ways to choose 2 items out of 10 is \(C(10,2)=\frac{10!}{2!(10 - 2)!}=\frac{10\times9}{2\times1} = 45\). The probability is \(\frac{C(7,2)}{C(10,2)}=\frac{21}{45}=\frac{7}{15}
eq\frac{49}{100}\).

Answer:

Chloe is incorrect because she assumed independence (as if picking with replacement) when using the \(10\times10\) grid. In reality, when picking two writing utensils together (without replacement), the sample - space should be calculated using combinations. The correct probability is \(\frac{7}{15}\), not \(\frac{49}{100}\).