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Thread: Conditional Probability and Independent Events Choosing a Team

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    Conditional Probability and Independent Events Choosing a Team

    A teacher randomly chooses a two-person leadership team from a group of four qualified students. Three of the students, Sandra, Marta, and Jane, are girls. The fourth student, Franklin, is a boy.


    Using the sample space of possible outcomes listed below, where each student is represented by the first letter of his or her name, answer each of the following questions.

    What is P(A), the probability that the first student is a girl?


    What is P(B), the probability that the second student is a girl?


    What is P(A and B), the probability that the first student is a girl and the second student is a girl?


    Are events P(A) and P(B) independent?

    The book states that these two events are independent. But the way the question is phrased, "chooses a two-person leadership team" makes me think they're dependent. If the Teacher chooses one girl there is one less girl in the population from which to choose. So P(A) = 3/4 but P(B) = 2/3.

    What is the difference between this and dealing a two-card hand where the probability of the first card being a 4 Spades and the second card being a 7 of Diamonds? Wouldn't that be P(A) = 1/52 and P(B) = 1/51?
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    Re: Conditional Probability and Independent Events Choosing a Team

    $P(B)$ is independent of the choice of the first student chosen. The first student chosen can be anyone. The second student chosen can be anyone left. Let's write up the sample space and we can decide which options yield positive results:
    Code:
    Outcome  (A)  (B)  (A and B)
    _______  ___  ___  _________
       SM    Yes  Yes     Yes
       SJ    Yes  Yes     Yes
       SF    Yes  No      No
       MS    Yes  Yes     Yes
       MJ    Yes  Yes     Yes
       MF    Yes  No      No
       JS    Yes  Yes     Yes
       JM    Yes  Yes     Yes
       JF    Yes  No      No
       FS    No   Yes     No
       FM    No   Yes     No
       FJ    No   Yes     No
    The first letter represents the first name of the first choice. The second letter represents the first name of the second choice. There are 12 possible outcomes, and among them, 9 have a girl as the first choice. Among them, 9 have a girl as the second choice. And among them, 6 have a girl as both the first and second choice.

    P(A) = 9/12, P(B) = 9/12, P(A and B) = 6/12
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    Re: Conditional Probability and Independent Events Choosing a Team

    Thanks SlipEternal. I knew the answers from the book. And I understand your explanation but it seems to me to defy reality.

    What I'm saying is that in the first choice there are 9 outcomes of the 12 where a girl is chosen. She's not available (unless you clone her) to be the second choice. So, whoever she is, there's only one of 2 other girls who can possibly be chosen out of a pool of 3 people. What's wrong with my logic?
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    Re: Conditional Probability and Independent Events Choosing a Team

    Quote Originally Posted by B9766 View Post
    Thanks SlipEternal. I knew the answers from the book. And I understand your explanation but it seems to me to defy reality.

    What I'm saying is that in the first choice there are 9 outcomes of the 12 where a girl is chosen. She's not available (unless you clone her) to be the second choice. So, whoever she is, there's only one of 2 other girls who can possibly be chosen out of a pool of 3 people. What's wrong with my logic?
    You are looking at conditional probability. If you assume you already chose a girl for the first slot, then find the probability of choosing a girl for the second slot. But that is not the probability of a girl being chosen for the second slot.

    Look at column (B) above. The last three choices have a boy chosen for the first position. If a boy is chosen first, there is a 100% chance the second slot goes to a girl. A and B are independent. You do not have to choose a girl first to choose a girl second. You can choose a boy first, then a girl second. Or you can choose two girls. The probability is:
    $P(\text{choose girl, then girl}) = \dfrac{3}{4}\cdot \dfrac{2}{3} = \dfrac{1}{2}$
    Plus
    $P(\text{choose boy, then girl}) = \dfrac{1}{4}\cdot \dfrac{3}{3} = \dfrac{1}{4}$
    for a total
    $P(\text{second slot goes to a girl}) = \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4}$
    Last edited by SlipEternal; Mar 8th 2018 at 08:03 AM.
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    Re: Conditional Probability and Independent Events Choosing a Team

    Thank you SlipEternal. I think I get it. And yes, I did assume the first girl was already chosen because of the sequence of the questions and the premise that a team of two was being built. Clearly a bad assumption on my part. I appreciate you taking the time to answer this.
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    Re: Conditional Probability and Independent Events Choosing a Team

    Quote Originally Posted by B9766 View Post
    A teacher randomly chooses a two-person leadership team from a group of four qualified students. Three of the students, Sandra, Marta, and Jane, are girls. The fourth student, Franklin, is a boy.
    What is P(A), the probability that the first student is a girl?
    What is P(B), the probability that the second student is a girl?
    What is P(A and B), the probability that the first student is a girl and the second student is a girl?
    Are events P(A) and P(B) independent?
    I am no way commenting on posts in this thread. It is just to old( I mean old)problem editor in me.
    A teacher randomly chooses a two-person leadership team, a chair & vice-chair, from a group of four qualified students. three girls and one boy.
    What is $\mathscr{P}(C_G)$, the probability that the chair is a girl?
    What is $\mathscr{P}(V_G)$, the probability that the vice-chair is a girl?
    Are events C_G and V_G independent events?

    Now note that we are now counting ordered pairs with no need to consider who is whom aside from sex.
    Look: $\mathscr{P}(C_G)=\frac{3}{4}$ so $\mathscr{P}(C_B)=\frac{1}{4}$
    But note that $\mathscr{P}(V_G)$ is conditioned upon the chair having been chosen male or female.

    Thus $ \begin{align*}\mathscr{P}(V_G)&=\mathscr{P}(V_G \cap C_G)+\mathscr{P}(V_G\cap C_B) \\&=\mathscr{P}(V_G | C_G)\mathscr{P}(C_G)+\mathscr{P}(V_G|C_B)\mathscr{ P}(C_B)\\&=\frac{2}{3}\frac{3}{4}+\frac{3}{3}\frac {1}{4}\\&=\frac{3}{4} \end{align*}$

    Question: Is it true that $\mathscr{P}(C_G | V_G)=\mathscr{P}(C_G )~?$ If yes does that mean independence?
    HOW & WHY?
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    Re: Conditional Probability and Independent Events Choosing a Team

    Quote Originally Posted by Plato View Post
    I am no way commenting on posts in this thread. It is just to old( I mean old)problem editor in me.
    A teacher randomly chooses a two-person leadership team, a chair & vice-chair, from a group of four qualified students. three girls and one boy.
    What is $\mathscr{P}(C_G)$, the probability that the chair is a girl?
    What is $\mathscr{P}(V_G)$, the probability that the vice-chair is a girl?
    Are events C_G and V_G independent events?

    Now note that we are now counting ordered pairs with no need to consider who is whom aside from sex.
    Look: $\mathscr{P}(C_G)=\frac{3}{4}$ so $\mathscr{P}(C_B)=\frac{1}{4}$
    But note that $\mathscr{P}(V_G)$ is conditioned upon the chair having been chosen male or female.

    Thus $ \begin{align*}\mathscr{P}(V_G)&=\mathscr{P}(V_G \cap C_G)+\mathscr{P}(V_G\cap C_B) \\&=\mathscr{P}(V_G | C_G)\mathscr{P}(C_G)+\mathscr{P}(V_G|C_B)\mathscr{ P}(C_B)\\&=\frac{2}{3}\frac{3}{4}+\frac{3}{3}\frac {1}{4}\\&=\frac{3}{4} \end{align*}$

    Question: Is it true that $\mathscr{P}(C_G | V_G)=\mathscr{P}(C_G )~?$ If yes does that mean independence?
    HOW & WHY?
    Plato,

    Two comments:
    1) I ran my original question by the Professor for the online course I'm taking. She said the problem was poorly written and could easily be taken as a conditional problem with dependency between P(A) and P(B). Logically, if one person is picked for a title, that person is not considered for any subsequent titles. By definition, there is no second thing if there wasn't a first. It's illogical to say, "Let us go to a bag of marbles and blindly select a second one and calculate the probability of that one being blue." So I feel somewhat relieved that I thought it illogical and the one grading my work agrees. I've found other questionable, and some downright wrong statements in the text and videos.

    2) I am just now starting the section on sets, intersections, unions, notation, and the Addition and Multiplication Rules so I'm having trouble following your logic. Please give me a few days to assimilate these ideas and then I'll take another stab at understanding what you've written. As always, I appreciate all your help.
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    Re: Conditional Probability and Independent Events Choosing a Team

    Quote Originally Posted by B9766 View Post
    "Let us go to a bag of marbles and blindly select a second one and calculate the probability of that one being blue." So I feel somewhat relieved that I thought it illogical and the one grading my work agrees
    I am just now starting the section on sets, intersections, unions, notation, and the Addition and Multiplication Rules so I'm having trouble following your logic.
    I far prefer to use bag problems to teach this. Suppose there are ten red balls and five blue balls in a bag. Draw two noting the color of the first and the the color of the second. What the probability that the second ball drawn is blue?
    Now here is the standard solution. The $\mathscr{P}{(B_2)}$, probability the second ball is blue, depends upon what color that the first ball drawn is. Well the first is either red or blue. Technicality that known as a partition of the space. So now we have: $B_2=(B_1\cap B_2)\cup(B_2\cap R_1)$ That says the second ball is blue happens in a space where the first ball is either red or blue.
    $ \begin{align*}\mathscr{P}{(B_2)}&=\mathscr{P}{(B_2 \cap B_1)}+\mathscr{P}{(B_2\cap R_1)}\\
    &=\mathscr{P}{(B_2|B_1)\cdot\mathscr {P}{(B_1)}}+\mathscr {P}{(B_2|R_1)\cdot \mathscr {P}{(R_1)}} \\ &=\frac{4}{15}\cdot\frac{5}{14}+\frac{5}{15} \cdot \frac{10}{14}\end{align*}$
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    Re: Conditional Probability and Independent Events Choosing a Team

    Plato, I don't understand your set notation. I understand intersections and unions but in what way is B1 a set? And in what way would B1 intersect with B2?
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    Re: Conditional Probability and Independent Events Choosing a Team

    Quote Originally Posted by B9766 View Post
    Plato, I don't understand your set notation. I understand intersections and unions but in what way is B1 a set? And in what way would B1 intersect with B2?
    $B_1\text{ nor }G_2$ is a set in the sense you are thinking. Rather they denote events.
    $B_1$ is the event the first ball is blue & $G_2$ is the event the second ball is green.
    Thus $B_1\cap G_2$ is the event the first ball is blue and(intersection) the second ball is green.

    $\mathscr{P}(B_1\cap G_2)=\mathscr{P}(G_2|B_1)\cdot\mathscr{P}(B_1)$ is standard notation:
    The probability first ball is blue and(intersection) the second ball is green is the probability the second ball is green given the probability first ball is blue times the probability first ball is blue .
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    Re: Conditional Probability and Independent Events Choosing a Team

    Ah! Thank you! Guess I need to learn the different notation for logic operations. This makes perfect sense now.
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    Re: Conditional Probability and Independent Events Choosing a Team

    Quote Originally Posted by B9766 View Post
    A teacher randomly chooses a two-person leadership team from a group of four qualified students. Three of the students, Sandra, Marta, and Jane, are girls. The fourth student, Franklin, is a boy.


    Using the sample space of possible outcomes listed below, where each student is represented by the first letter of his or her name, answer each of the following questions.

    What is P(A), the probability that the first student is a girl?
    That is, of course, 3/4.


    What is P(B), the probability that the second student is a girl?
    This is a little ambiguous. Are we to assume that the first person chosen was a girl? Since this question itself does not say so, I would be inclined to ignore the first question and say that, given three girls and one boy the probability the second person chosen is a girl is 3/4. That is different from "if the first person chosen is a girl, what is the probability the second person chosen is also a girl. If the first person chose is a girl, that leaves two girls and one boy so the probability the second person is also a girl is 2/3.

    Notice that we could also do this by arguing that if the first person chosen is a girl, which has probability 3/4, then the probability the second person chosen is also a girl is 2/3 while if the first person chosen is a boy, which has probability 1/4, then the probability the second person chosen is a girl is 3/3= 1. The 'a-priori' probability that the second person chosen is a girl, without regard to what gender the first person drawn is, is (3/4)(2/3)+ (1/4)(1)= 2/4+ 1/4= 3/4 as above.


    What is P(A and B), the probability that the first student is a girl and the second student is a girl?
    The probability the first person chosen is a girl is 3/4 and, given that, the probability the second person chosen is a girl is 2/3. The probability both people chosen are girls is (3/4)(2/3)= 1/2.


    Are events P(A) and P(B) independent?
    Again, that depends upon interpretation. The second problem only asked for P(B), the probability that the second person chosen was a girl. Iit made no reference to the first problem, the first student being a girl. Interpreted that way, the two are independent.

    The book states that these two events are independent. But the way the question is phrased, "chooses a two-person leadership team" makes me think they're dependent. If the Teacher chooses one girl there is one less girl in the population from which to choose. So P(A) = 3/4 but P(B) = 2/3.

    What is the difference between this and dealing a two-card hand where the probability of the first card being a 4 Spades and the second card being a 7 of Diamonds? Wouldn't that be P(A) = 1/52 and P(B) = 1/51?
    If you were given a problem that asked for P(B), the probability that, out of 52 cards, the second card drawn is the 7 of diamonds, without reference to what the first card drawn was, then P(B)= 1/52, not 1/51.
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