Lucky Contest Winners

The problem below is the first selection by the NCTM President Mike Shaughnessy for the newly introduced section of their Newsletter Problem to Ponder.

Students at your school have just finished competing in the qualifying round of a nationally sponsored contest on mathematical reasoning and sense making. When the work was scored, it turned out that four students at your school all had perfect preliminary papers-two girls and two boys. The school decided to hold a random drawing among these four students to select two of them to send to the national finals. The drawing takes place in the school auditorium. You show up late to the drawing, just as one of the winners-a girl-is leaving the stage amid cheers.

1. Suppose that the girl that you saw leaving the stage is the first winner. What is the probability that the second winner will also be a girl?

2. Suppose that the girl that you saw leaving the stage was the second winner. What is the probability that the first winner was also a girl?

Solution

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Copyright © 1996-2017 Alexander Bogomolny

Students at your school have just finished competing in the qualifying round of a nationally sponsored contest on mathematical reasoning and sense making. When the work was scored, it turned out that four students at your school all had perfect preliminary papers-two girls and two boys. The school decided to hold a random drawing among these four students to select two of them to send to the national finals. The drawing takes place in the school auditorium. You show up late to the drawing, just as one of the winners-a girl-is leaving the stage amid cheers.

1. Suppose that the girl that you saw leaving the stage is the first winner. What is the probability that the second winner will also be a girl?

2. Suppose that the girl that you saw leaving the stage was the second winner. What is the probability that the first winner was also a girl?

The simplest way to approach the problem is to consider the associated sample space. There are two girls and two boys from which a pair of winners is going to be drawn. Let's denote the girls G and g and the boys B and b. Formally speaking, there are 16 (= 4×4) possible ordered pairs formed by the four letters. However, from the context of the problem, the repetitions, like GG, make no sense and should be excluded outright.

For the completeness sake I added one more problem:

Students at your school have just finished competing in the qualifying round of a nationally sponsored contest on mathematical reasoning and sense making. When the work was scored, it turned out that four students at your school all had perfect preliminary papers-two girls and two boys. The school decided to hold a random drawing among these four students to select two of them to send to the national finals. The drawing takes place in the school auditorium. You pass a door to the auditorium, peek in, see a girl descending the stage, but, being pressed in time, close the door and continue on your errand.

3. You do not know whether the girl that you saw leaving the stage was the first or the second winner. What is the probability that the other winner is also a girl?

The three problems (with the last coming first) are collected in the tables below.

  the sample space for the Lucky Contest Winners problem

The legend is this: the impossible pairs (two boys or the same girl twice) are marked with a '×'. The two girl combinations are marked with '+' and all the others with '-'. In the first question (the middle table), the first member of the pair is known to be a girl, making the four combinations in the lower left corner impossible. These are also colored blue. A similar treatment applies to the upper right corner in the last table (second question.) Of the remaining squares the ones with favorable outcomes - Gg or gG - are colored yellow, the unfavorable ones red.

For the first two questions there are 6 possible outcomes: 2 favorable and 4 unfavorable. In both cases then, the probability that two girls have been selected to advance to the national finals is 2/6 = 1/3. For the third problem, there are 10 possible outcomes, with only two favorable. The probability of having two lucky girls is then 2/10 = 1/5.

Solution 3

In the July 15, 2010 issue of the Newsletter, we have two solutions:

  1. If the girl that you saw on the stage was the first winner, then out of the three students left for the random drawing for a second winner, only one is a girl, and two are boys. The chances that the second one is also a girl will be 1 in 3, or 1/3.

    However, if the girl that you saw on stage was actually the second winner, then the chance that the first winner was also a girl is 1/2, because before the first girl was picked, there were two girls and two boys, so the chance of a choosing a girl was then 2 in 4, or 1/2.

  2. (Same as above) If the girl that you saw on the stage was the first winner, then out of the three remaining students, one is a girl and two are boys, so the chances that the second winner will also be a girl is 1/3.

    (Different than above) It doesn't make any difference whether you saw the first girl or the second girl; the fact that you saw a girl winning at all means that there is only a 1 in 3 chance that the other winner was a girl, so, the probability that the first winner was also a girl if the girl that you see on the stage is the second winner is also 1/3.

    The solutions are followed by an advice to run a simulation. Before you do, give a thought to the fact that the second part of the first solution does not exploit the fact that the you saw a girl as the second winner. The argument would be the same if you saw a boy. This should make one suspicious of this line of reasoning.

    To strengthen the second (and mine) solution we may apply the Bayes' theorem. In our notations, we would like to evaluate P(G|g) - the probability that the first winner was a girl provided the second winner was also a girl. We accept as known (obvious or from the argument above) that

    • P(G) = P(B) = 1/2, i.e., the probabilities of the first winner being a girl or a boy are both equal to 1/2.

    • P(g|G) = 1/3, i.e., the probability of the second winner being a girl provided the first one was a girl is 1/3 (My solution and one of the NCTM's).

    • P(g|B) = 2/3, i.e., the probability of the second winner being a girl provided the first one was a boy is 2/3. This is because when the first boy was out after the first drawing, there remain two girls and one boy for the second drawing.

    Now let's use Bayes' formula:

     P(G|g)= P(Gg) / P(g)
      = P(g|G)P(G) / (P(g|G)P(G) + P(g|B)P(B))
      = 1/3·1/2 / (1/3·1/2 + 2/3·1/2)
      = 1/3 / (1/3 + 2/3)
      = 1/3.

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    Lokking back, it might have been easier to evaluate P(Gg) outright. This the probability that the two winners were both girls. There are C(4, 2) = 6 to chose 2 items out of four. We are only interested in one selection. Therefore, P(Gg) = 1/6. Divided by P(g) which is 1/2, we obtain P(G|g) = 1/6 ÷ 1/2 = 1/3.

    Copyright © 1996-2017 Alexander Bogomolny

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