Page 30 - DMTH404_STATISTICS

Statistics

Notes We now give a number of examples to show how to calculate the probabilities of events in a
variety of situations. Please go through these examples carefully. If you understand them. you
will have no difficulty in doing the exercises later.

Example 8: A box contains ninety good and ten defective screws. Let us find the probability
that 5 screws selected at random out of this box are all good.
Let A be the event that the 5 selected screws are all good.
Now we can choose 5 screws out of 100 screws in ways. If the selected 5 screws are to be good,
they will have to be selected out of the 90 good screws. This can be done in ways. This is the
number of sample points favourable to A. Hence the probability of A

Example 9: A government prints 10 lakh lottery tickets of value of Rs. 2 each. We would
like to know the number of tickets that must be bought to have a chance of 0.5 or more to win the
first prize of 2 lakhs.
6
The prize-winning ticket can be mndomly selected out of the 10 lakh tickets in 10 ways.
Now, let m denote the number of tickets that we must buy. Then m is the number of points
favourable to our winning the first prize. Therefore, the probability of our winning the first
m
prize, is, 6 .
10

m 1 10 6 10 6
Since we want that  , therefore m  . This means that we must buy at least =
10 6 2 2 2
500,000 tickets, at a cost of at least Rs. 10 lakhs ! Not a profitable proposition at all !

Example 10: In a study centre batch of 100 students, 54 opted for MTE-06,69 opted for
MTE - 11 and 35 opted for both MTE-06 and MTE-11. If one of these students is selected at random,
let us find the probability that the student has opted for MTE-06 or MTE- 11.
Let M denote the event that the randomly selected student has opted for MTE-06 and S the event
that helshe has opted for MTE- 11. We want to know P(M U S). According to the classical
54 69 35
definition. P(M) = , P(S) = and P(M  S) = . Thus
100 100 100
P(M S) = P(M) + P(S) – P(MS)
Suppose now we want to know the probability that the randomly selected student has opted for
C
C
neither MTE-06 nor MTE-11. This means that we want to know P[M  S ].
Now,
C
C
M  S = (M  S) C
Therefore,
C
C
P(M S ) = 1 – P[M  S] = 1 – 0.88 = 0.12
Lastly, to obtain the probability that the student has opted for MTE-06 but not for MTE-11, i.e., to
C
C
C
obtain P(MS ), observe that M = (MS)(MS ) and that MS and MS are disjoint
events. Thus.
P(M) = P(M  S) + P(M  SC)

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