Page 13 - Life Insurance Today March 2016
P. 13
d. Here the solution of the Example is - there are 13 (Probability that Y dies in 2 years) X
balls in all, of which 5 are Green. The probability (Probability that Z survives 2 years) = .002 X
that the first ball is Green is 5/13. Having drawn .003 X (1 - .004) = .000006 X .996 =
the Green ball first, there are 8 Black balls and 4 .000005976.
Green balls. So P2, the probability that the second
ball is Black, after a Green ball having been drawn (ii) Example B: A coin is thrown twice. What is the
already in the first trial, is 8/12. Therefore, the probability that head turns up at least once?
required probability in this case is 5/13 X 8/12 = 40/
156. Solution here, is - this event "at least one
head in two throws" is made up of the
e. Let us take another Example : With data of the following mutually exclusive events are -
above Example find out the probability that one
of the balls drawn is Black and the other Green (a) Just one head in two throws;
without specifying the order.
(b) Head in both the throws.
f. Here the solution of Example is - One Black ball
and one Green ball can be drawn in the following The first of these is again made up of (i) head in
two mutually exclusive ways: the 1st throw and tail in the 2nd throw, and (ii)
tail in the 1st throw and head in the 2nd throw.
(i) First ball is Green and the second ball is Black; Thus the mutually exclusive events making up the
desired event are:
(ii) First ball is Black and the second ball is Green.
(i) Head at 1st throw and tail at 2nd throw;
g. If we can find the probabilities for (i) and (ii)
separately, we can add the result and finally get (ii) Tail at 1st throw and head at 2nd throw;
the answer.
(iii) Head in both the throws.
h. We have seen that the probability for (i) above is
5/13 X 8/12 = 40/156. As for the 2nd event, viz. The probability of (i) = (Probability of head at the
first ball being Black and the second ball being 1st throw) X (Probability of tail at the 2nd throw)
Green, we observe that the probability that the = ½ X ½ = 1/4, as the two component events are
first ball drawn is Black is 8/13. Having drawn a independent.
Black ball, there are 12 balls of which 5 are Green
and so probability of drawing a Green ball after The probability of (ii) = (Probability of tail at the
drawing a Black in the first trial is 5/12. Hence the 1st throw) X (Probability of head at the 2nd
probability required under (ii) is 8/13 X 5/12 = 40/ throw) = ½ X ½ = 1/4, as the two component
156. So finally here the required answer = 40/156 events are independent
+ 40/156 = 80/156.
The probability of (iii) = (Probability of head at the
i. Now let us consider how these concepts are 1st throw) X (Probability of head at the 2nd
utilized in Life Insurance calculations in next throw) = ½ X ½ = 1/4, as the two component
three examples- events are independent
(i) Example A: Of three persons X, Y and Z, the Therefore, the required probability is obtained by
probability that R dies within 2 years is .002, adding the probability of these three mutually
Y dies within two years is .003 and Z dies exclusive events and is thus equal to ¼ + ¼ + ¼ =
within 2 years is .004. What is the probability ¾.
that both R and Y die during the period of two
years while Z survives two years? Alternatively, instead of finding the required
probability directly we may first find the
Solution here, is - Required Probability = probability of the complementary event. The
(Probability that R dies in 2 years) X complementary event of "at least one head" is "no
heads at all" i.e. tail in 1st throw followed by tail
in 2nd throw.
"Without music, life would be a mistake."
Life Insurance Today March 2016 13
