Probability Question

While the probability question can be very straightforward, you need to pay attention to the detail of what you are being asked. I gave this question (1999) in the mini-mock exam and nobody got it right.

In your exam, you can answer questions like this using the sum or product of probabilities as appropriate or using a sample space. While the sample space might take some time to set up, it is a foolproof method. To illustrate how the probabilities work, I have created the matching sample space for each part of the question. (you will need to click on the image to be able to read the text within the sample space diagram)

Note that there are 66 possible combinations, as you cant pick the same exact sample twice and selecting e.g. A1 and B2 is the same as selecting B2 and A1.


Looking at part (i):From the sample space you can count the 10 cases where this can happen. So the probability is 10/66 = 5/33


Using probabilities only, you work it out like this:

P(1st is type A) = 5/12

P(2^nd is type A) = 4/11 [If the first one was an A, then there will only be 4 As left. Also, there will only be 11 samples left, as the first sample isn't replaced.]

P(1st is type A AND 2^nd is type A) = 5/12 x 4/11 = 20/132 = 5/33

Looking at part (ii):

Counting from the sample space you get 12/66 = 2/11.
Working this out using probabilities is trickier.
Here you have to remember that there are two ways this could happen. The first way is that you get a B first, then an O, OR the second is the other way around. In both cases you have to remember that you are choosing from 12 the first time you pick and from 11 the second time.
The probabilities work out like this:
P(B then O OR O then B) :
Work out P(B then O) first:
P(1st is type B) = 4/12
P(2nd is type O) = 3/11
So P(B AND O) = 4/12 x 3/11 = 12/132
Then work out P(O then B)
P(1st is type O) = 3/12
P(2nd is type B) = 4/11
So P(O AND B) = 3/12 x 4/11 = 12/132
(Note you could have assumed that you "B followed by O" is equally likely as "O followed by B" so that probabilities would be the same).
Now work out P(B then O OR O then B)
= P(B then O) + P(O then B)
= 12/132 + 12/132
= 24/132
= 12/66 (same as sample space above)
Simplify to 2/11.

Looking at part (iii):From the sample space you can see that this can occur 19 ways out of 66 so P(both the same type) = 19/66.
Using probabilities, you need to consider what is a successful outcome. If you write out what "of the same blood type" means, you will see that a successful outcome is "both A OR both B OR both O", so you need to add P(both A) + P(both B) + P(both O).
You have already found the probabilty of both samples being type A in part (i).
The answer was 10/66 or 5/33.
Now you need to find the probability of both being B and the probability of both being O.
P(Both B) = 4/12 x 3/11 = 12/132 = 6/66
P(Both O) = 3/12 x 2/11 = 6/132 = 3/66
(I am keeping all fractions with 66 as common denominator on purpose).
Now add
P(Both A OR both B OR both O) = 10/66 + 6/66 + 3/33 = 19/66