2019 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 17 Medium

A red ball and a green ball are randomly and independently tossed into bins numbered with positive integers so that for each ball, the probability that it is tossed into bin k is 2^{-k} for k=1, 2, 3, \cdots What is the probability that the red ball is tossed into a higher-numbered bin than the green ball? (2019 AMC 10B Problem, Question#17)

  • A.

    \frac{1}{4}

  • B.

    \frac{2}{7}

  • C.

    \frac{1}{3}

  • D.

    \frac{3}{8}

  • E.

    \frac{3}{7}

Answer:C

By symmetry, the probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher-numbered bin.Clearly, the probability of both landing in the same bin is \sum _{k=1}^{\infty}2^{-k}\cdot 2^{-k}= \sum _{k=1}^{\infty}2^{-2k}= \frac{1}{3} (by the geometric series sumformua). Therefore the other two probabilites have to both be \frac{1- \dfrac{1}{3}}{2}= \text {(C)}\frac{1}{3}.

Suppose the green ball goes in bin i, for some i\geqslant 1. The probability of this occuring is \frac {1}{2^i}. Given that this occurs, the probability that the red ball goes in a higher-numbered bin is \frac{1}{2^{i+1}}+ \frac{1}{2^{i+2}}+ \cdots = \frac{1}{2^{i}} (by the geometric series sum formula). Thus the probability that the green ball goes in bin i, and the red ball goes in a bin greater than i, is (\frac{1}{2^{i}})^{2}= \frac{1}{4^{i}}. Summing from i=1 to infinity, we get \sum_{i=1}^{\infty}\frac{1}{4^{i}}=\text {(C)}\frac{1}{3}. where we again used the geometric series sum formula. (Alternatively, if this sum equals n, then by writing out the terms and multiplying both sides by4, we see 4n=n+1, which give n= \frac{1}{3}).

The probability that the two balls will go into adjacent bins is \frac{1}{2 \times 4}+ \frac{1}{4 \times 8}+ \frac{1}{8 \times 16}+ \cdots = \frac{1}{8}+ \frac{1}{32}+ \frac{1}{128}+ \cdots = \frac{1}{6} by the geometric series sum formula. Similarly, the probability that the two balls will go into bins that have a distance of 2 from each other is \frac{1}{2 \times 8}+ \frac{1}{4 \times 16}+ \frac{1}{8 \times 32}+ \cdots = \frac{1}{16}+ \frac{1}{64}+ \frac{1}{256}+ \cdots = \frac{1}{12} (again recognizing a geometric series). We can see that each time we add a bin between the two balls. the probability halves. Thus, our answer is \frac{1}{6}+ \frac{1}{12}+ \frac{1}{24}+\cdots, which, by the geometric series sum formula， is \text {(C)}\frac{1}{3}.

Define a win as a ball appearing in higher numbered box.

Start from the first box.

There are 4 possible results in the box: Red, Green, Red and Green, or none with an equal probablly of 4 for each. if none of the balls is in the first box, the game restarts at the second box with the same kind of probaility distribution, so if p is the probability that Red wins, we we can write p= \frac{1}{4}+ \frac{1}{4}p; there is a \frac 14 probabilty that "Red” wins immediately, a 0 probability in the cases ”Geen” or “Red and Green”, and in the ”None" case (which occurs with \frac{1}{4} probability), we then start again, giving the same probability p. Hence,solving the equation, we get p= \text {(C)}\frac{1}{3}.

Write out the infinite geometric series as \frac12, \frac14,\frac18, \frac{1}{16}, \cdots To find the probablilty that red goes in a higher-numbered bin than green, we can simply remove all odd-index terms(i.e term 1, term 3, etc.)， and then sum the remaining terms-this is in fact precisely equivalent to the method of Solution 2.

Writing this out as another infinite geometric sequence, we are left with \frac{1}{4}, \frac{1}{16}, \frac{1}{64},\cdots Summing, we \sum_{get i=1}^{\infty}\frac{1}{4^{i}}=\text {(C)}\frac{1}{3}.

Related Problems
Problem 15 2019 AMC 10 B
Problem 16 2019 AMC 10 B
Problem 18 2019 AMC 10 B
Problem 19 2019 AMC 10 B
Think Academy Powered by ChatGPT