# Thread: Conditional Probability and Independent Events Choosing a Team

1. ## 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?

2. ## 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

3. ## 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?

4. ## Re: Conditional Probability and Independent Events Choosing a Team

Originally Posted by B9766
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
$P(\text{second slot goes to a girl}) = \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4}$

5. ## 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.

6. ## Re: Conditional Probability and Independent Events Choosing a Team

Originally Posted by B9766
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?

7. ## Re: Conditional Probability and Independent Events Choosing a Team

Originally Posted by Plato
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,

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*}