Page 14 - Life Insurance Today March 2016

The probability for both tails = (Probability of tail a person aged 45 survives 5 years) X (Probability that
in 1st throw X Probability of tail in 2nd throw) =
½ X ½ = ¼. Now the required probability can
definitely be obtained by subtracting the above
probability from 1 and is thus equal to 1 - ¼ = ¾
(as calculated before in other method).
(iii) Example C: The probability that a person aged
40 dies in five years is .04, that a person aged
45 dies in five years is .06 and that a person
aged 50 dies in five years is, 08. Find the
probability that of the three persons aged 40,
45 and 50 respectively in the following cases
- (7) Theory of Permutations: It facilitates to decide the
(a) When exactly one survives 5 years;
(b) When at least one survives 5 years;
(c) When at least one dies in 5 years;
(d) When the person aged 40 dies between
ages 50 and 55.
Solution: (a) The event "exactly one survives five
years" can be broken up into the following three
mutually exclusive events;
(i) A (aged 40) survives 5 years while B (aged 45) and C
(aged 50) both die within 5 years.
(ii) B survives 5 years while both A and C die within 5
years.
(iii) C survives 5 years while both A and B die within 5
years.
The probability for (i) = .96 X .06 X .08 = .004608.
The probability for (ii) = .94 X .04 X .08 = .003008.
The probability for (i) = .92 X .04 X .06 = .002208.
Finally the required probability = .004608 + .003008
+ .002208 = .009824.
Solution: (b) The complementary event of "at least
one survives five years" is "none survives 5 years" i.e.
A, B and C all die within 5 years. Probability for this
situation = .04 X .06 X .08 = .000192. Therefore, the
probability that when at least one survives 5 years is
= 1 - .000192 = .999808.
Solution: (c) The required probability = 1- Probability
that all the three survive 5 years = 1 - (Probability that
a person aged 40 survives 5 years) X (Probability that
"Insanity is doing the same thing, over and over again, but expecting different results."
14 March 2016
a person aged 50 survives 5 years) = 1- (1- Probability
that a person aged 40 dies in 5 years) X (1- Probability
that a person aged 45 dies in 5 years) X (1- Probability
that a person aged 50 dies in 5 years) = 1- (1-.04) (1-
.06) (1-.08) = 1-.830208 = .169792.
Solution: (d) The required probability = (Probability
that a person aged 40 survives 5 years) X (Probability
that a person aged 45 survives 5 years) X (Probability
that a person aged 50 dies within 5 years) = .96 X .94
X .08 = .072192.
total number of equally,likely,mutually,exclusive and
exhaustive ways in which an event can happen and
the number of ways favourable for happening of the
event in a particular desired way. The theorem states
- If event E1 can happen in n1 ways, event E2 can
happen in n2 ways, even E3 can happen in n3 ways,
and so on, independent of each other, then all the
events E1, E2, E3 ….. together or in succession can
happen in n1 X n2 X n3..… number of ways.
We have seen that when two dice are thrown there
are 36 equally,likely,mutually, exclusive and exhaustive
ways in which the numbers appear on upper faces. If
E1 be the event of the number appearing on upper
face of one dice and E2 that of the other dice, E1 can
happen in 6 ways and E2 can also happen in 6 ways.
Thus E1 and E2 can happen in 6 X 6 = 36 ways
together if the two dice are thrown simultaneously;
or in succession if the dice are thrown one after the
other. Permutations are used - Say, the number of
ways in which r objects out of the given n (unlike)
objects can be arranged (in a straight line) is defined
as permutations of N objects taken R at a time. The
first object can be selected in n ways as any one of
the n objects can be chosen. After selecting the first
object, the 2nd object can be selected in (n-1) ways
out of the remaining (n-1) objects. Similarly the 3rd
object can be selected in (n-2) ways and so on. The
rth object can be selected in n - (r - 1) i.e. (n - r+1)
ways.
So number of ways in which r objects can be selected
= n (n-1) (n-2) ….. (n - r+1).
Now let us consider the permutations of n objects of
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