Page 12 - Life Insurance Today March 2016
P. 12
1, 2 or 3 then P1 = ½. Let event E2 be defined to of the number 1, 2, 3, 4, 5 or 6 can appear on 2nd
occur when it turns up 2, 4 or 6 then P2 =1/2 - Dice. The same argument holds good for other
here E1 and E2 are not mutually exclusive because number appearing on 1st dice.
when 2 turns up both E1 and E2 occur
simultaneously. The probability of either E1 or E2 c. So representing the above events or incidents in
to occur is the probability that the dice will turn a tabular form we have the following :
up 1,2,3,4 or 6 i.e. 5/6 which is not equal to P1+P2
(i.e. 1). No. appearing Numbers appearing Total no.
on 1st Dice on 2nd Dice of ways
For example: A card is drawn from a pack of 52
cards. The probability that the card drawn is - 1 1 or 2 or 3 or 4 or 5 or 6 6
2 1 or 2 or 3 or 4 or 5 or 6 6
i. An Ace is 1/13 (ii) a king is 1/13 (iii) a Queen
is 1/13 and (iv) also a jack is 1/13. It may thus 3 1 or 2 or 3 or 4 or 5 or 6 6
be concluded that drawing of Ace or King or
Queen or Jack are mutually exclusive events. 4 1 or 2 or 3 or 4 or 5 or 6 6
Therefore, the probability that a card drawn
is either an Ace, or a King, or a Queen or a 5 1 or 2 or 3 or 4 or 5 or 6 6
Jack is = 1/13+1/13+1/13+1/13= 4/13.
6 1 or 2 or 3 or 4 or 5 or 6 6
(5) It follows the Multiplication Theorem of Probability-
It can be explained by the following example: Total: - 36 ways
Example: If E1 and E2 are two independent events d. Thus there are total 36 number of equally,likely,
which occur with probability P1 and P2 respectively. mutually, exclusive and exhaustive ways out of
Then the probability that both the events E1 and E2 which only 1 way is possible of the given events.
happen together is the product of two probabilities Therefore, the desired probability is 1/36 as
i.e. P1X P2. shown above.
a. It is important to note that the events should be
e. The application of the theorem simplifies the
independent i.e. happening of one event should working, and complicated probabilities can be
not influence the happening of the other. worked out conveniently.
Consider that when two dice are thrown, then the
probability that the number '3' appears on the 1st (6) Multiplication Theorem of Probability which needs to
Dice and number '4' on the second Dice. The be taken in General Form is detailed out below:
probability that the 1st dice will turn up No. 3 is
P1 = 1/6 and similarly, the probability that 2nd The multiplication theorem mentioned above is a
dice will turn up No. 4 is P2= 1/6. Also here, the special case (being applicable to independent events
number appearing on 1st dice in no way only) of a more general theorem which is described
influences the number appearing on the 2nd dice. in seriatim below:
Therefore, the two events are independent. a. If P1 is the probability of happening of an event
Hence, the probability that '3' appears on the first
dice and '4' appears on the 2nd dice is given by E1 and P2 is the probability of another event E2
P1x P2 = 1/6 X 1/6 = 1/36. happening after the event E1 has happened, then
the probability that both the events E1 and E2
b. This can be verified by considering the number of happening in succession is given by P1 X P2,
favourable ways and total number of equally,likely, whether E1 and E2 are independent or not.
mutually,exclusive and exhaustive ways. When 1st
dice is used - any of the number 1, 2, 3, 4, 5 or 6 b. If E1, E2 are independent, P2 is simply the
may appear. Again when 1 appears on 1st dice any probability that E2 happens as it is not influenced
by the fact whether E1 has happened or not.
c. Let us consider this Example : A bag contains 8
Black balls and 5 Green balls. Two balls are drawn
one at a time without replacing the ball drawn.
What is the probability that the first ball is Green
and second one is Black?
"To live is the rarest thing in the world. Most people exist, that is all."
12 March 2016 Life Insurance Today
