1. ## Probability Question

Hi, I have been stuck with this example for quite a while now. The question is as follows:

A customer entering a bicycle store will buy a new bike with a probability of 10% and will buy bicycle accessories with a probability of 65%. With a probability of 30% a customer will not purchase anything.

(a) Find the probability that a customer will buy both a new bike and bicycle accessories.
and (b) Find the probability that a customer will only buy a bike.

I have done similar questions before however with this question I don't quite understand what to do with.

For part (a) I did the following:

A = buy a bike. P(A) = 0.1
B = buy bike accessories. P(B) = 0.65

Then I tried to use the formula:
P(A u B) = P(A) + P(B) - P(A n B) but I don't know how I would find P(A n B).

In the end I got a value of 0.685 but don't think its correct.

Thanks

2. Originally Posted by axa121
Hi, I have been stuck with this example for quite a while now. The question is as follows:

A customer enters a bicycle store and will buy a new bike with a probability of 10% and will buy bicycle accessories with a probability of 65%. With a probability of 30% that a customer will not buy anything.

(a) Find the probability that a customer will buy a new bike and bicycle accessories.
and (b) Find the probability that a customer will only buy a bike.

I have done similar questions before however with this question I don't quite understand what to do with.

For part (a) I did the following:

A = buy a bike. P(A) = 0.1
B = buy bike accessories. P(B) = 0.65

Then I tried to use the formula:
P(A u B) = P(A) + P(B) - P(A n B) but I don't know how I would find P(A n B).

In the end I got a value of 0.685 but don't think its correct.

Thanks
Start by drawing a Karnaugh table that represents the given information:

$\begin{tabular}{l | c | c | c} & B & B' & \\ \hline A & & & 0.1 \\ A' & & 0.3 & \\ \hline & 0.65 & & 1 \\ \end{tabular}
$

Fill in the gaps.

Then get the required probability from the table.

3. This is what I got for the table:

A -------B-----------B'

A-------0.05------0.05--------0.1

A'-------0.6---------0.3-------0.9

---------0.65------0.35---------1

4. Oh maybe I was interpreting it in a different way, as I was thinking that the B itself has a probability of 0.65.

It's been a while since I've done stats, and the way I was reading it, I thought you would n't use the P(B|A) formula.

I've used that now and got the value for P(B n A) as 0.065

Is it correct?

edit: I've also put the actual wording of the question up now.

5. Originally Posted by axa121
Oh maybe I was interpreting it in a different way, as I was thinking that the B itself has a probability of 0.65.

It's been a while since I've done stats, and the way I was reading it, I thought you would n't use the P(B|A) formula.

I've used that now and got the value for P(B n A) as 0.065

Is it correct? Mr F says: Is 0.065 the number in the cell of the Karnaugh table coresponding to $\Pr(A \cap B)$?

edit: I've also put the actual wording of the question up now.
..

6. No it is not.