Example.1
An unbiased coin is tossed. If the result
is a head, a pair of unbiased dice is rolled and the number obtained by adding
the numbers on the two faces is noted. If the result is a tail, a card from a
wellshuffled pack of eleven cards numbered 2, 3, 4,L,
12 is picked and the number on the card is noted. Find the probability that
noted number is either 7 or 8.
Solution
Let A_{1}(A_{2}) be the
event that tossing of coin results in head (tail).
And A be the event that noted number
is 7 or 8.
P(A_{1}) = P(A_{2})
=
P(A/A_{2}) =
P(A/A_{1}) =
(As 7 can result in, (1, 6), (2,5), (3,4),
(6,1), (5,2), (4, 3) i.e. 6 ways and 8 can result in (2, 6), (3,5), (4,4),
(6,2), (5,3) i.e. 5 ways).
Now P(A) = P(A_{1})×P(A/A_{1}) + P(A_{2})×P(A/A_{2})
= =
Example.2
In a multiplechoice question there are
four alternative answers, of which one or more than one are correct. A
candidate will get marks on the question only if he ticks all the correct
answers. The candidate decides to tick answers randomly. If he is allowed up to
three chances to answer the questions, find the probability that he will get
marks on it.
Solution
Total number of ways of answering the
question,
= ^{4}C_{1}
+ ^{4}C_{2} + ^{4}C_{3} + ^{4}C_{4}
= 15
Now out of these,
only one combination is correct.
&⇒ Probability of answering the
question correctly at the first trial =
Now, Probability of
answering the question correctly at the second trial =
(with the assumption
that student doesn’t repeat the mistake of first trial)
Similarly,
probability of answering the question correctly at the third trial
=
&⇒ Probability of required event =
Example.3
Sixteen players S_{1},
S_{2},...,S_{16} play in a tournament. They are divided into
eight pairs at random. From each pair a winner is decided on the basis of a
game played between the two players of the pair. Assume that all the players
are of equal strength .
(a) find the probability that the player S_{1}
is among the eight winner
(b)
find the probability that exactly one of the two players S_{1} and S_{2}
is among the eight winner. .
Solution
(a) Since all the players are of
equal strength and eight players are winners. So probability that S_{1}
will be among the winner equals to 8/16 = 1/2.
(b) Let A_{1} be the event
that S_{1}, S_{2} play in same pair and A_{2 }be the
event that they play in different pair and A be the event that exactly one out
of S_{1} and S_{2} is among the winners.
P(A) = P(A_{1})×P(A/A_{1})
+ P(A_{2})×P(A/A_{2})
Now P(A_{1}) =
P(A_{2}) = 1

Therefore, P(A) = .
Example.4
A manufacturing company uses an
acceptance scheme on production items before they are shipped. The plan is a
twostage one. Boxes of 25 items are readied for shipment and a sample of 3 are
tested for defectives. If any defectives are found, the entire box is sent for
100% screening. If no defectives are found, the box is shipped.
(a) what is the probability that a box
containing 3 defectives will be shipped?
(b) what is the probability that a box
containing only one defective will be sent back for screening?
Solution
(a) A box containing
3 defectives will be shipped, if no defective item is included in the sample.
&⇒ Required probability =
(b) In this case
sample taken must contain the defective item,
&⇒ Required probability =
Example.5
One of ten keys
opens the door. If we try the keys one after another, what is the probability
that the door is opened on the (i) first attempt, (ii) second attempt, (iii)
tenth attempt.
Solution
Since out of 10 keys
only one can open the door so door will open in first attempt with probability Now if he fails in first attempt
which has the probability then he will
attempt next time with different key. So probability that door will open in
second attempt is
Probability that
door will open only in tenth attempt is equal to the probability that door not
opened in first nine attempts which is equal to (As
each time he tries with new key and keeps away the key which does not open the
door).