Page 31 - DMTH404_STATISTICS

Unit 2: Methods of Enumeration of Probability

or P(M  SC) = P(M) – P(M  S) Notes

54 35
=   0.19.
100 100

Example 11: A throws six unbiased dice and wins if he has at least one six. B throws
twelve unbiased dice and wins if he has at least two sixes. Who do you think is more likely to
win?
We would urge you- to make a guess first and then go through the following computations.
Check if your intuition was correct.
6
12
The total number of outcomes for A is n = 6 and that for B is n = 6 , We will first calculate the
A B
probabilities qA and qB that A and B, respectively, loose their games. Then the I probabilities of
their winning are P = 1 – q and P = 1 – q , respectively. We do this because q and q are easier
A A B B A B
to compute.
Now A loses if he does not have a six on any of the 6 dice he rolls. This can happen in 5 different
6
6
ways, since he can have nu sis on each die in 5 ways. Hence q = 5 /6 and therefore, P =
6
A A
1 – (5/6)  0.665.
6
In order to calculate q , observe that B loses if he has no six or exactly one six. The probability
B
that he has no six is 5 /6 = (5/6) . Now the single six can occur on any one of the 12 dice, i.e.,
12
12
12
 12 
in   ways. Then all the remaining 11 dice have to have a score other than six. This can happen
 1 
11
in 5 ways.
 12  11
Therefore, the total number of ways of obtaining one six is   5 . Hence the probability that
 1 
12 5 11

B has exactly one six is 12 .
6
The events of “no six” and “one six” in the throwing of 12 dice are disjoint events. Hence the
probability
5 11
12
q = (5/6)  12  0.381
B 12
6
Thus, P  1 – 0.38 1 = 0.619.
B
Comparing PA and PB, we can conclude that A has a greater probability of winning.
Now here are some exercises which you should try to solve.
So far we have seen various examples of assigning probabilities to sample points and have also
discussed some properties of probabilities of events. In the next section we shall talk about the
concept of conditional probability.

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