# Probability Test

• Apr 22nd 2008, 11:36 PM
andrew2322
Probability Test
Gah i posted 2 questions about my test earlier, cud u pls check it
• Apr 23rd 2008, 12:44 AM
Moo
Hello,

Questions 1, 2 and 3 are ok.

Question 4 :

Quote:

a.) The first marble picked is black and the second is red: 5/36
I don't see any black marble :/

Quote:

b.) Obtaining 2 white marbles: 1/16
Ok.
• Apr 23rd 2008, 12:48 AM
andrew2322
sorry
Find One Lol
• Apr 23rd 2008, 12:52 AM
Moo
Good job ! (Clapping)
• Apr 23rd 2008, 01:01 AM
andrew2322
hey
booo
• Apr 23rd 2008, 01:03 AM
Moo
Yes ^^

Unless I made mistakes too, but all the reasoning is correct !
• Apr 23rd 2008, 01:05 AM
andrew2322
hey
haha i doubt two people can make the exact same mistake
• Apr 23rd 2008, 04:55 AM
colby2152
So far #1 and #2 are correct, but #3 has an error...

Quote:

3.) A red and a black die labelled 1 - 6 are thrown simultaneously, find the probability of:

a.) A number less than 5 on the red die: 2/3
b.) A number less than 5 on the red die but a totla of 10 altogether: 1/36
c.) Both die adding to 10: 1/12
a) is correct

given that, b) should be equal to $\frac{2}{3}*\frac{1}{6} = \frac{1}{9}$

c) is correct, three possibilities out of thirty-six

Quote:

4.) There is 3 white marbles, 4 blue marbles and 5 red marbles in a bag. Find the probabilty of if marbles are picked WITH REPLACEMENT:

a.) The first marble picked is black and the second is red: 5/36
b.) Obtaining 2 white marbles: 1/16
a) 0, there are no black marbles
b) $\frac{3}{12}^2 = \frac{9}{144}$, Remember, the white marble is replaced by another white marble.

#5, #6 and #7 are all correct

Note: Best way to solve #7 is by using a Venn diagram
• Apr 23rd 2008, 09:09 AM
Moo
Quote:

Originally Posted by colby2152
a) 0, there are no black marbles

He said it was blue

Quote:

b) $\frac{3}{12}^2 = \frac{9}{144}$, Remember, the white marble is replaced by another white marble.
The probability to get a white marble is 3/12=1/4
Since there is replacement, the probability to get 2 marbles will be the product of the probabilities of getting 1 marble, that is to say 1/4²=1/16, isn't it ?
Actually, 9/144 is 1/16... :p
• Apr 23rd 2008, 09:11 AM
Moo
Quote:

Originally Posted by colby2152
given that, b) should be equal to $\frac{2}{3}*\frac{1}{6} = \frac{1}{9}$

The thing is that the red dice has to be < 5, so 1 2 3 or 4.
Since the total has to be 10, the only possibility is that the red dice is 4 and the blue dice is 6 -> 1/36
• Apr 25th 2008, 12:01 PM
colby2152
Quote:

Originally Posted by Moo
The probability to get a white marble is 3/12=1/4
Since there is replacement, the probability to get 2 marbles will be the product of the probabilities of getting 1 marble, that is to say 1/4²=1/16, isn't it ?
Actually, 9/144 is 1/16... :p

Yep, $\frac{9}{144} = \frac{1}{16}$

Quote:

Originally Posted by Moo
The thing is that the red dice has to be < 5, so 1 2 3 or 4.
Since the total has to be 10, the only possibility is that the red dice is 4 and the blue dice is 6 -> 1/36

Both adding to 10 is the event of a 6 on one die and a 4 on another. This probability is 1/36. I read the question wrong.