conditional probability

• Nov 10th 2007, 11:40 AM
valosn
conditional probability
I'm having trouble solving a problem. I've been staring at it for almost an hour now, and have had hardly any progress...

I'm trying to do problems 1 and 4 from here.

• Nov 10th 2007, 02:28 PM
Soroban
Hello, valosn!

Here's #1 . . .

Quote:

1. A factory produces items in boxes of two.
Over the long run:
. . (A): 92% of the boxes contain 0 defective items,
. . (B): 5% of the boxes contain 1 defective item, and
. . (C): 3% of the boxes contain 2 defective items.

A box is picked at random from production, then an item is picked
at random from that box. Given that the item is defective, what is
the probability that the second item in the box is defective?

We are concerned with two events:
. . $1D$: the first item is defective, and
. . $2D$: the second item is defective.

We want: . ${\color{blue}P(2D\,|\,1D) \;=\;\frac{P(1D \wedge 2D)}{P(1D)}}$ .[1]

The numerator is: $P(1D \wedge 2D)$, the probability that both are defective ... case (C).
. . We are told that this happens: $3\%$ of the time.
Hence: . ${\color{blue}P(1D \wedge 2D) \:=\:\frac{3}{100}}$

The denominator is: $P(1D)$, the probability that the first is defective.
This can happen in two ways:

(i) Case (B) which happens $\frac{5}{100}$ of the time
. . and we pick the defective item, probability $\frac{1}{2}$.
Thus, this has probability: $\frac{5}{100}\cdot\frac{1}{2} \:=\:\frac{1}{40}$

(ii) Case (C) which happens $\frac{3}{100}$ of the time
. . and we pick a defective item, probability $1.$
Thus, this has probability: $\frac{3}{100}(1) \:=\:\frac{3}{100}$

Hence: . $P(1D) \:=\:\frac{1}{40} + \frac{3}{100} \quad\Rightarrow\quad{\color{blue}P(1D) \:=\:\frac{11}{200}}$

Substitute into [1]: . $P(2D\,|\,1D) \;=\;\frac{\dfrac{3}{100}}{\dfrac{11}{200}} \;=\;\boxed{\frac{6}{11}}$

• Nov 10th 2007, 03:21 PM
Jhevon
Here's 4

recall that we say two events are independent if knowledge that one occurs does not change the probability of the other occurring.

(a) here, there are different numbers of each type of ball in both boxes. however, the numbers are proportional. no matter which box we select, the probability of selecting a black ball from that box is the same, and the probability of selecting a red ball from that box is the same. thus, yes, the color of the ball is independent of which box is chosen. verify this using the formulas for conditional probability and show that you get the same result for selecting, say, a black ball given that you have selected box 1, and then, the probability of selecting a black ball given that you have selected box 2. do this for the red ball also

(b) the events are now dependent. knowledge of which box we've selected changes the probability of a black ball being chosen. verify this using the formulas.
• Nov 10th 2007, 03:28 PM
valosn
Thanks a lot! It was the denominator that was giving me trouble, but you explained it very clearly. Thank you!

As for the 4th problem, I was having trouble with part b), but I figured it out.

Again, thanks a lot for the help!
• Nov 10th 2007, 03:50 PM
Plato
I have that book in front of me. I reviewed an early edition for Springer.
For #4, first find P(R).
$\begin{array}{rcl}
P(R) & = & P(R \cap 1) + P(R \cap 2) \\
& = & P(R|1)P(1) + P(R|2)P(2) \\
& = & \frac{2}{5}\frac{1}{2} + \frac{8}{{20}}\frac{1}{2} \\
& = & \frac{2}{5} \\ \end{array}$

Now we know that
$P(R|1) = \frac{2}{5}\frac{1}{2} = \frac{1}{5}\,\& \,P(R|2) = \frac{8}{{20}}\frac{1}{2} = \frac{1}{5}$

Do you think they are independent?