1. ## Probability Problem

1.) There are 3 white marbles, 4 blue marbles and 5 red marbles in a bag. If they are picked at random with replacement, find the probability of:

a.) The first being a blue and the second a red. I got: 5/36

b.) The probability they are both whites. I got 1/16

and this one:

If Pr(A) = 0.5
Pr(B) = 0.4
and Pr(A and B) = 0.1

Find Pr(A' and B') i got: 0.2
Find Pr(A' and B) i got: 0.3
Find Pr(A or B) i got: 0.8

THANKS!

2. a)
the probability that you get a blue then a red is:

$\frac{4}{12} \times \frac{5}{12} = \frac{15}{144}$

both whites are:

$\frac{3}{12} \times \frac{3}{12} = \frac{9}{144}$

and for b)

I'm assuming A' means the complement of A? so we get

$P(A^c \cap B^c) = 1-P(A \cap B) = 1-0.1 = 0.9$

$P(A^c \cap B) = 0.4-0.1 = 0.3$

$P(A \cup B) = 0.5+0.4-0.1 = 0.8$

3. Originally Posted by andrew2322

1.) There are 3 white marbles, 4 blue marbles and 5 red marbles in a bag. If they are picked at random with replacement, find the probability of:

a.) The first being a blue and the second a red. I got: 5/36

b.) The probability they are both whites. I got 1/16

and this one:

If Pr(A) = 0.5
Pr(B) = 0.4
and Pr(A and B) = 0.1

Find Pr(A' and B') i got: 0.2
Find Pr(A' and B) i got: 0.3
Find Pr(A or B) i got: 0.8

THANKS!

(Lucky you boldfaced the with replacement )

4. Originally Posted by lllll
a)
the probability that you get a blue then a red is:

$\frac{4}{12} \times \frac{5}{12} = \frac{15}{144}$ Mr F says: *Ahem* That's 20/144 (= 5/36).

both whites are:

$\frac{3}{12} \times \frac{3}{12} = \frac{9}{144}$ Mr F says: = 1/16.

and for b)

I'm assuming A' means the complement of A? so we get

$P(A^c \cap B^c) = 1-P(A \cup B) = 1-(0.5+0.4-0.1) = 0.2$

$P(A^c \cap B) = 0.4-0.1 = 0.3$

$P(A \cup B) = 0.5+0.4-0.1 = 0.8$
By the way, for b) a Karnaugh table is a fast and efficient way of doing this problem. You'll find a few examples in this forum. For example: http://www.mathhelpforum.com/math-he...3-prob-qn.html

5. ## hey thanks guys

Originally Posted by lllll
a)
the probability that you get a blue then a red is:

$\frac{4}{12} \times \frac{5}{12} = \frac{15}{144}$

both whites are:

$\frac{3}{12} \times \frac{3}{12} = \frac{9}{144}$

and for b)

I'm assuming A' means the complement of A? so we get

$P(A^c \cap B^c) = 1-P(A \cap B) = 1-0.1 = 0.9$

$P(A^c \cap B) = 0.4-0.1 = 0.3$

$P(A \cup B) = 0.5+0.4-0.1 = 0.8$
so which one is right? Mr.F or IIII lol im so confused :P

6. Originally Posted by andrew2322
so which one is right? Mr.F or IIII lol im so confused :P
(4)(5) is 20, not 15. So clearly lllll made a small arithmetic slip up in part a.