Sequences: Consecutive Integers - Average=Median

A bag contains one chip labeled "1," two chips labeled "2," three chips labeled "3," and so on, all the way up to 100 chips being labeled "100." If one chip is randomly chosen, then what is the chance that it has an even number less than or equal to 50?

Incorrect. [[snippet]] This answer is probably the result of summing the even numbers from 2 to 100 for the numerator instead of using only the even numbers from 2 to 50.

Incorrect. [[snippet]] You might have gotten this answer if you summed all the numbers from 1 to 50 in the numerator instead of using just the even numbers.

Incorrect. [[snippet]] Even though the even numbers less than 50 amount to one-fourth of all the integers from 1 to 100, there are a different number of chips for each—for example, there's only one "1" chip, while there are one hundred "100" chips). This needs to be taken into account when calculating the probability.

Now, find the denominator. For this, you need to find the total number of chips, which is equal to $$1+2+3+...+100$$. You can calculate it using the same method you used for the numerator. The average of this sequence is $$\hspace{0.5in} \displaystyle \mbox{Average} = \frac{1+100}{2} = 50.5$$. The number of terms is $$\hspace{0.5in} \displaystyle \mbox{Number of Terms} = (100-1)+1 = 100$$. Thus, the sum (i.e., the total number of outcomes) is $$\hspace{0.5in} \mbox{Sum} = 50.5 \cdot 100 = 5050$$. Finally, divide the numerator by the denominator to find the probability. $$\hspace{0.5in} \displaystyle P(E) = \frac{650}{5050} = \frac{13}{101}$$

Incorrect. [[snippet]] You might have gotten this answer choice if you used the sum of the even numbers from 2 to 48 for the numerator. However, the problem actually asks for the probability of choosing an even number less than or equal to 50.

That's right! The probability formula is $$\hspace{0.5in} \displaystyle P(E)=\frac{\mbox{Number of Outcomes in }E}{\mbox{Total Number of Outcomes}}$$. You're trying to calculate the chance of picking an even number less than or equal to 50, so for the top of the fraction, you need to calculate the number of even chips from 2 to 50. Since there are two chips labeled "2," four chips labeled "4," and so on, there are a total of $$2+4+6+...+50$$ even chips labeled from 2 to 50. You can calculate this sum by multiplying its average by the number of terms in it. Since it's a sequence of consecutive even integers, its average is equal to the average of the extremes. $$\hspace{0.5in} \displaystyle \mbox{Average} = \frac{2+50}{2} = 26$$ You can calculate the number of terms by subtracting the extremes, dividing by 2, and then adding 1. $$\hspace{0.5in} \displaystyle \mbox{Number of Terms} = \frac{50-2}{2} + 1 = 25$$ Finally, multiply these numbers to find the sum (i.e., the number of outcomes in _E_). $$\hspace{0.5in} \mbox{Sum} = 26 \cdot 25 = 650$$

$$\displaystyle \frac{12}{101}$$

$$\displaystyle \frac{13}{101}$$

$$\displaystyle \frac{1}{4}$$

$$\displaystyle \frac{51}{202}$$

$$\displaystyle \frac{51}{101}$$

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