# Thread: A Probability Problem

1. ## A Probability Problem

The Problem: Here is a game you might play. An urn contains seven red and three green balls. You are to select a ball, note its colour, and then replace it. Your oponent us then to select a ball. Ry is the event that you select red:Ro us the event that your oponent selects red;Gy is the event that you select green, etc. You win if the ball your opponent selects has the same colour as the ball you selected. What is the probability of winning? Make a tree diagram of possible outcomes.

Attempted Solution:

Two branches

Branch I leads to Ry
Probability 7/10

Branch 2 leads to Gy
Probability 3/10

Branch I branches in two directions

Branch III leads to Ro
Probability 6/9

Branch IV leads to Go
Probability 3/9

Branch II branches in two directions

Branch V leads to Go
Probability 2/9

Branch VI leads to Ro
Probability 7/9

To find the probability of winning, I simply add the probabilities of each event which leads to a win.

The probability of you getting two reds is (7/10)(6/9).
The probability of you getting two greens is (3/10)(2/9).
Both cases lead to a win.

So I added the two cases.

P(winning) = (7/10)(6/9) + (3/10)(2/9)
= 42/90 + 6/90
= 48/90

But, so the key informs me, the answer is 29/50. How could this be?

2. *Ahem* Note what I have redded and bolded ....

Originally Posted by D. Martin
The Problem: Here is a game you might play. An urn contains seven red and three green balls. You are to select a ball, note its colour, and then replace it. Your oponent us then to select a ball. Ry is the event that you select red:Ro us the event that your oponent selects red;Gy is the event that you select green, etc. You win if the ball your opponent selects has the same colour as the ball you selected. What is the probability of winning? Make a tree diagram of possible outcomes.

Attempted Solution:

Two branches

Branch I leads to Ry
Probability 7/10

Branch 2 leads to Gy
Probability 3/10

Branch I branches in two directions

Branch III leads to Ro
Probability 6/9

Branch IV leads to Go
Probability 3/9

Branch II branches in two directions

Branch V leads to Go
Probability 2/9

Branch VI leads to Ro
Probability 7/9

To find the probability of winning, I simply add the probabilities of each event which leads to a win.

The probability of you getting two reds is (7/10)(6/9).
The probability of you getting two greens is (3/10)(2/9).
Both cases lead to a win.

So I added the two cases.

P(winning) = (7/10)(6/9) + (3/10)(2/9)
= 42/90 + 6/90
= 48/90

But, so the key informs me, the answer is 29/50. How could this be?
(7/10)(7/10) + (3/10)(3/10) = 49/100 + 9/100 = 29/50.

3. D. Martin!

An urn contains seven red and three green balls.
You are to select a ball, note its colour, and then replace it.
Your oponent is then to select a ball.
$R_y$: you select red. . $G_y$: you select green.
$R_o$: opponent selects red. . $G_o$: opponent selects green.
You win if the ball your opponent selects has the same colour as the ball you selected.

What is the probability of winning?
Make a tree diagram of possible outcomes.

. . . $\begin{array}{cccccccccccc}
& & & & & * \\
& & & & / & & \backslash \\
& & & ^{\frac{y}{10}}_/ & & & & _\backslash^{\frac{3}{10}} \\
& & R_y & & & & & & G_y \\
& ^{\frac{7}{10}}_/ & & _{\backslash}^{\frac{3}{10}} & & & & ^{\frac{7}{10}}_/ & & _{\backslash}^{\frac{3}{10}} \\
R_o & & & & G_o & & R_o & & & & G_o \\ \\[-3mm]
\frac{49}{100} & & & & \frac{21}{100} & & \frac{21}{100} & & & & \frac{9}{100}\end{array}$

$P(\text{win}) \;=\;\frac{49}{100} + \frac{9}{100} \;=\;\frac{58}{100} \;=\;\frac{29}{50}$