# Thread: Probability Problem (tree diagram)

1. ## Probability Problem (tree diagram)

A couple has three babies.
a. what are the chances 3 are girls?
b. what are the chances there will be boy,girl,boy?

http://i49.tinypic.com/2isyo0.jpg

^ is tree diagram

a.

n(S)=(2^3) = 8
n(A)=7?

P(A)=7/8

b. n(S)=(2^3)=8
n(A)=3

P(A)=3/8

Also.. another from the textbook:

Suppose your school's basketball team is playing a four-game series against another school. So far this season, each team has won three of the six games in which they faced each other.

a) Determine the probability of your school winning exactly two games. (use your tree diagram)

I'm not sure how to do this.

n(S)=(2^4)=16
n(A) = ?

P(A) = n(A)/16

How to find n(A) here i'm not sure

2. Hello Johnboe
Originally Posted by Johnboe
A couple has three babies.
a. what are the chances 3 are girls?
b. what are the chances there will be boy,girl,boy?

http://i49.tinypic.com/2isyo0.jpg

^ is tree diagram

a.

n(S)=(2^3) = 8
n(A)=7?

P(A)=7/8

b. n(S)=(2^3)=8
n(A)=3

P(A)=3/8

Also.. another from the textbook:

Suppose your school's basketball team is playing a four-game series against another school. So far this season, each team has won three of the six games in which they faced each other.

a) Determine the probability of your school winning exactly two games. (use your tree diagram)

I'm not sure how to do this.

n(S)=(2^4)=16
n(A) = ?

P(A) = n(A)/16

How to find n(A) here i'm not sure
Question 1

Your tree diagram is correct. But in (a) only one of the $8$ possible outcomes is favourable ( $3$ girls). So the probability of this is $\tfrac18$.

In (b), if the question means B, G, B in that order, then, again, only one outcome is favourable. So again the probability is $\tfrac18$. However, if the question means "What is the probability that they have 2 boys and a girl?", then your answer $\tfrac38$ is correct.

Question 2

Of course, in reality, the results of previous games cannot be used with any degree of certainty to predict future results. However, if we assume that, based on past form, the probability that your team wins a particular game is $\tfrac12$, then we can draw a tree diagram to represent the $16$ possible sets of outcomes of four games, with each outcome equally likely. (Just extend the tree diagram you had in Question 1 to another level.)

You'll find that, of the $16$ possible paths through the tree, $6$ of them result in $2$ wins and $2$ losses. So the probability of exactly $2$ wins is $\tfrac{6}{16}= \tfrac38$.

(If you're not sure what these $6$ possible paths are, here they are: WWLL, WLWL, WLLW, LWWL, LWLW, LLWW.)