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1MATHEMATICS Chapter 12: Probability
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1. CONDITIONAL PROBABILITYSuppose we toss two fair dice, and
note the sum of the numbers on the two dice. What is the
probability that the sumis 9. The answer is clear,
36, 4,5 , 5, 4 , 6,3 , 3,6n S E i.e., n (E) = 4
4 1
36 9P E
Now, if we put a condition to the above problem, that the first
die shows a 3. What will be probability of getting a sumof 9?In
this case the sample space will not be the same.
' 4,5 , 5, 4 , 6,3 , 3,6S
' 3,6E
1'4
E
When number 3 has already occured, we can see how the
probability changes. This is conditional probability.The
conditional probability of an event B in relationship to an event A
is the probability that event B occurs given thatevent A has
already occurred. The notation for conditional probability is
P(B|A), read as the probability of B given A.
general formula
A BAB B
PP
P
In other words P(A/B) = Probability of occurence of A when B is
taken as the sample space
A BNumber of elementaryevents favourable to A BP A / B
Number of elementaryevents favourable to B Bn
n
Let S be the sample space containing both the events A and
B.
Therefore ,
A BS P A B
P A / BB P BS
nnnn
We can understand this general formula with the help of the
following reasoning:If the event B occurs, then in order for A to
occur, it is necessary that the actual occurrence be a point in
both A and Bi.e., it must be in A B . And as B has already
occurred, B becomes the new sample space.Properties of conditional
probabilityProperty 1If A and B are any two events of a sample
space S and F is an event of S such that 0P F , then
/ / / /P A B F P A F P B F P A B F
If A and B are disjoint events, then / / /P A B F P A F P B F
Property 2
/ 1 /P E F P E F
Proof: Since, / 1P E E F
By using property 1, we get / / / /P E E F P E F P E F P E E
F
Since E and E are disjoint events / 0P E E F , we get
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1 / /P E F P E F
Or / 1 /P E F P E F
Properties 1S FP PF F
Proof: Let E and F be events of a sample space S of an
experiment, then we have* Note: (b) If A and B are two events such
that , 0,A B PB then
(i) P(A/B) =
P A
P A B P AP B
(ii) /P A B P A (iii) P(B/A) = 1
We know that
P (S | F) =
P S F P F1
P F P F
AlsoP (F | F) =
P F F P F1
P F P F
Thus P (S | F) = P (F | F) = 1 Probabilitity of occurrence of an
event when it has already occured is 1
*Note : (a) If A and B are two events such that P(A/B) = P(B/A)
, then P(A) = P(B).
Proof :
A ; /B
P A B P A BP P B A
P B P A
P A P B
2. MULTIPLICATION THEOREM ON PROBABILITYLet A and B be two
events associated with a sample space S. We already know that the
conditional probability of eventA given that B has occurred is
given by
P A B
P A / B , P B 0P B
Above can be written as P A / B .P B P A B 1 Also, we know that
the conditional probability of event B given that A has occurred is
given by
P A B
P B / A , P A 0P A
Or P B / A .P A P A B 2 By combining (1) and (2), we get
P A B P A / B .P B P B / A .P A Which is multiplication rule of
probability.Multiplication Rule of probability for more than two
eventsIf A, B and C are three events of sample space, then we can
write
P A B C P A .P B /A .P C / A B 3. INDEPENDENT EVENTS
Two events, A and B, are independent if the occurrence of A does
not affect the probability of occurring of B.Examples of
independent events are: Tossing a coin for Head and picking a card
for an ace from a deck of 52 cards Throwing a die for even number
to appear and picking a ball from a urn
When two events, A and B, are independent, the probability of
both occurring is P A B P A P B
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For three independent events A, B and C, we have P A B C = P A
.P B .P C Take two independents events A and B. Because occurrence
of A does not depend on the occurrence of B.
P(A/B) = P(A) and P(B/A) = P(B)By multiplication rule of
probability
/ .P A B P A B P B
.P A B P A P B Note: Events can also be mutually exclusive. Two
events are said to be mutually exclusive if occurrence of one
eliminates all
chances of the occurrence of other. For e.g. E be the event that
a child is present for the class on a particular day and Fbe the
event that a child is absent for the class on that day. Now if E
occurs, F cannot occur i.e., both events cannotoccur
simultaneously.
E F (If A and B take test. Event of A solving a question is
independent of the event of B solving a question (provided theyare
not cheating!)
For independent events, . .P E F P E P F P E P F P E F P E P
F
For mutually exclusive events, P E F P E P F P E F
4. BAYES THEOREMLet S be a sample space and if A1, A, A3 ... An
are mutually exclusive and exhaustive events such that
1 2 3A A A ......A Sn
Then for any event A which is a subset of S, we have
1
P E P A/EP E / A
P E P A/E
i ii n
j jj
for any i = 1,2,3,.n
Proof: By Conditional probability, we can write
P A E
P E / AP A
ii
Or
P E P A/E
P E / AP Ai i
i
By using theorem of total probability, we get
1
P E P A/EP E / A
P E P A/E
i ii n
j jj
.
Law of total probability should be before Bayes Theorem.
Theorem (Law of Total Probability) If E1, E2, E3, .., En are
mutually exclusive and exhaustive events of the sample
space S, then for any event A of S.
1 1 2 2 3 3P A P E .P A/E P E .P A/E P E .P A/E ...... P E .P
A/En n 1
P E .P A/En
i ii
Proof: We have from the Venn diagram
1 2A= E A E A ..... E An 1 2 ......................n are
mutually exclusive events
Now, 1 2P A =P E A P E A ..... P E A 1n We already know by
conditional probability that
P E A P E P A/Ei i i Hence equation (1) can be written as 1 1 2
2P A =P E P A/E P E P A/E ..... P E P A/En n
1
P A P E P A/En
i ii
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5. RANDOM VARIABLEA random variable is a real valued function
whose domain is the sample space of a random experiment.A random
variable can take ay real value. Thus, its co-domain is the set of
real number.Discrete Random VariableA random variable which can
assume only a finite number of values or countably infinite values
is called a discreterandom variable.For a random experiment of
tossing two coins simultaneously. Let X denotes the number of tails
then, X is a randomvariable which can take values 0, 1,
2.Continuous random variableA random variable which can assume all
possible values between certain limits is called a continuous
random variableDiscrete Probability DistributionA discrete random
variable assumes each of its values with a certain probability. Let
X be a discrete random variablewhich takes values x1, x2, x3,xn
where pi = P{X = xi}Then X : x1 x2 x3 .. xn P(X): p1 p2 p3 ..... pn
is called the probability distribution of x,
where, 1
0, 1, 1, 2,......n
i ii
p p i n
Mean and Variance of a Discrete Random VariableLet X be a
discrete random variable which can assume values x1, x2, x3,xn with
probabilities p1, p2, p3 .. pnrespectively then. Mean of X denoted
by E(X) or is given by
1 1 2 21
E X .......n
i i n ni
x p x p x p x p
Variance of X denoted by 2 is given by
221
n
i ii
x p
or 2 2 21
n
i ii
x P
2
2 2
1 1
n n
i i i ii i
x P x p
or, 22 2 E Ex x
222 2 2
1 1
E X E Xn n
i i i ii i
x p x p
Where 2 21
n
i iq
E X x p
Standard Derivation :
The non-negative number 21
Var Xn
i ii
x p
is called the standard deviation of the random variableX.
6. BERNOULLI TRIALS AND BINOMIAL DISTRIBUTION
Trials of a random experiment are called Bernoulli trials, if
they satisfy the following conditions:
1. Each trial has only two possible outcomes, generally called
success and failure.2. The trials are independent. Intuitively, the
outcome of one trial has no influence over the outcome of another
trial.3. On each trial, the probability of success is p and the
probability of failure is 1 - p.4. There should be finite number of
trials, i.e., the probability of success & failure remains the
same in each trial.Note: when n > 3, it is easier to solve the
problem with the help of B.D.
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Following are some Bernoulli Trials Experiments
Flip a coin. Take a penalty shot on goal. Test a randomly
selected circuit to see whether it is defective. Roll a die and
determine whether it is a 6 or not.A binomial random variable is a
random variable that counts the number of successes in a sequence
ofindependent Bernoulli trials with fixed probability of
success.Binomial DistributionThe probability of achieving exactly k
successes in n trials is shown below.
P successesin n trials n k n kkk C p q Where,n = number of
trialsk = number of successesn k = number of failuresp =
probability of success in one trialq = 1 p = probability of failure
in one trialThus, the probability distribution of numbers of
success in an experiment consisting of n Bernoulli trials may
be
obtained by the binomial expansion of nq p . Hence, this
distribution of number of success X can be written as
we know (q + p) n = 1 0
1n
n nc q p
Note : (1) Mean = np
(2) Variance = n pq
(3) Standard deviation = variance n pq
X 0 1 2 3 --- k --- n P(x)
0n nC q 1 11
n nC q p 2 22n nC q p 3 33
n nC q p --- n n k kkC q p --- n nnC p
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20111. A random variable X has the following probability
distribution.
X 0 1 2 3 4 5 6 7 P(X) 0 K 2K 2K 3K K2 2K2 7K2 + K
Determine(i) K (ii) P(X < 3) (iii) P(X > 6) (iv) P (0 <
X < 3)
OrFind the probability of throwing at most 2 sixes in 6 throws
of a single die.
2. Given three identical boxes I, II and III each containing two
coins. In box I, both coins are gold coints, in box II, bothare
silver coins and in box III, there is one gold and one silver coin.
A persion chooses a box at random and takes outa coin. If the coin
is of gold, what is the probability that the other coin in the box
is also of gold ?
20101. On a multiple choice examination with three possible
answers (out of which only one is correct) for each of the five
questions, what is the probability that a candidate would get
four or more correct answers just by guessing.[Ans. 11/243] 4
Marks
2. A card from a pack of 52 cards is lost. From the remaining
cards of the pack, two cards are drawn at random and arefound to
both clubs. Find the probability of the lost card being of clubs. 6
Marks
[Ans. 11/50]3. From a lot of 10 bulbs, which includes 3
defectives, a sample of 2 bulbs is drawn at random. Find the
probability
distribution of the number of defective bulbs.
[Ans.
X 0 1 2 F(x) 7/15 7/15 1/15
] 6 Marks2009
1. Two cards are drawn simultaneously (or successively without
replacement) from a well shuffled pack of 52 cards. Findthe mean
and variance of the number of red cards. [Ans. 1, 25/51] 6
Marks
2. The probability that A hits a target is 1/3 and the
probability that B hits it is 2/5. If each one of A and B shoots at
thetarget, what is the probability that (i) the target is hit. (ii)
exactly one of them hits the target. [Ans. 1/5, 1/3]
4 Marks2008
1. A pair of dice is thrown 4 times. If getting a doublet is
considered a success, find the probability distribution of numberof
successes. 4 Marks
[Ans.
X a 1 2 3 4
F(x) (5/6)4 9C135
6
13
4C235
6
21
6
4C3 56
316
4
44
1C
6
]
2. An insurance company insured 2,000 scooter drivers, 4,000 car
drivers and 6,000 truck drivers. The probability of anaccident
involving a scooter, a car and a truck are 0.01, 0.03 and 0.15
respectively. One of the insured persons meetswith an accident.
What is the probability that he is a scooter driver? [Ans.
1/103]
6 Marks2007
1. An urn contains 7 red and 4 blue balls. Two balls are drawn
at random with replacement. Find the probability of getting
(a) 2 red balls (b) 2 blue balls (c) one red and one blue ball.
[Ans. 7 4 7 4
2 2 1 111 11 11
2 2 2
, ,C C C CC C C ] 2 Marks
2. A card is drawn at random from a well-shuffled pack of 52
cards. Find the probability that it is neither an ace nor a
king.[Ans. 11/13] 2 Marks
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7MATHEMATICS Chapter 12: Probability
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3. There are two bags I and II. Bag I contains 2 white and 3 red
balls and Bag II contains 4 white and 5 red balls. One ballis drawn
at random from one of the bags and is found to be red. Find the
probability that it was drawn from bag II.
[Ans. 25/53] 5 Marks
4. Find mean variance 2 for the following probability
distribution: 5 MarksX 0 1 2 3
P(X) 18
38
38
18
[Ans. 2
3 3,2 4
]Or
Find the binomial distribution for which the mean is 4 and
variance 3. [Ans. 1 316, ,4 4
A p q ]2006
1. Two dice are rolled once. Find the probability that: 3
Marks(i) the numbers on the two dice are different [Ans. 25/36](ii)
the total of numbers on the two dice is at least 4 [Ans. 11/12]
2. A pair of dice is tossed twice. If the random variable X is
defined as the number of doublets, find the probability
distribution of X. [Ans.
X 0 1 2 F(x) 25/36 5/18 1/36 ] 3 Marks
3. In a factory, which manufactures nuts, machines A, B and C
manufacture respectively 25%, 35% and 40% of nuts. Oftheir output
5, 4 and 2 per cent respectively are defective nuts. A nut is drawn
at random from the product and is foundto be defective. Find the
probability that it is manufactured by machine B. [Ans. 28/69]
3 Marks4. If the mean and variance of the Binomial distribution
are respectively 9 and 6, find the distribution. 3 Marks
[Ans. A = 27, P= 1/3, q = 2/3]2005
1. A class consists of 10 boys and 8 girls. Three students are
selected at random. Find the probability that the selectedgroup has
3 Marks(a) All boys (b) All girls (c) 2 boys and 1 girl. [Ans.
5/34, 7/102, 15,34]
2. A company has two plants to manufacture motor cycle. 70%
motor cycles are manufactured at the first plant, while 30%are
manufactured at the second plant. At the first plant, 80% motor
cycles are rated of the standard quality while at thesecond plant,
90% are rated of standard quality. A motor cycle, randomly picked
up, is found to be of standard quality. Find the probability that
it has come out from the second plant. [Ans. 56/83] 4 Marks
3. The probability that a student entering a university will
graduate is 0.4. Find the probability that out of 3 students of
theuniversity: 4 MarksNonewillgraduate, Onlyonewillgraduate,
Allwillgraduate. [Ans. 0.216, 0.432, 0.064]
20041. An urn contains 7 white, 5 black and 3 red balls. Two
balls are drawn at random. Find the probability that
(i) Both the balls are red [Ans. 1/21] 3 Marks(ii) One ball is
red, the other is black [Ans. 1/7](iii) One ball is white [Ans.
8/15]
2. Three urns A, B and C contain 6 red and 4 white: 2 red and 6
white: and 1 red and 5 white balls respectively. An urn ischosen at
random and a ball is drawn. If the ball drawn is found to be red,
find the probability that the ball was drawnfrom urn A. [Ans.
36/61] 3 Marks
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PRACTICE QUESTIONS1. A bag contains 6 red and 5 blue balls and
another bag contains 5 red and 8 blue balls. A ball is drawn from
the first bag
and without noting its colour is put in the second bag. A ball
is then drawn from the second bag. Find the probability
that the ball drawn is blue in colour. [Ans. 93
154 ]
2. A box contains 5 green and 4 yellow marbles. The second box
contains 3 green and 4 yellow marbles. Third boxcontains 4 green
and 4 yellow marbles. One marble is drawn from one of the three
boxes. What is the probability that the
marble which is drawn is yellow? [Ans. 191378 ]
3. Urn I has 2 white and 3 black balls, urn II has 4 white and 1
black ball and urn III has 3 white and 4 black balls. An urnis
selected at random and a ball is drawn at random.(a) What is the
probability of drawing a white ball?
(b) If the ball drawn is white, what is the probability that urn
I was selected? [Ans. (i) 1935 , (ii)
1457 ]
4. In a factory, a product is manufactured by any of three
machines A, B and C. They produce respectively 25%, 35% and40% of
the total products. A product is selected at random. Find the
probability that(i) it is defective (ii) it is produced by the
machine A, B or C given that it is defective.Assume that machines
A, B and C produce respectively 5%, 4% and 2% defective items.
[Ans. (i) 0.0345, (ii) 0.36, 0.41, 0.23]5. It is known that 40%
of the students in a certain college are girls and 50% of the
students are above the median height.
If 2/3 of the boys are above the median height, what is the
probability that a randomly selected student who is below
median height is girl? [Hint. First calculate that 75% girls are
below median and 13 of boys are below median]
[Ans. 0.6]6. A scooter manufacturing company has two plants.
Plant I manufactures 70% of the scooters and plant II
manufactures
30%. At plant I, 80% of the scooters are of standard quality and
at plant II, 90% of scooters are rated of standard quality.A
scooter is chosen at random and is found to be of standard quality.
What is the probability that it has come from plantII ?
[Ans. 2783 ]
7. An insurance company insured 1500 scooter drivers, 2500 car
drivers and 4500 truck drivers. The probability of ascooter, a car
and a truck driver meeting with an accident is 0.01, 0.02 and 0.04
respectively. If one of the insured person
meets with an accident, find the probability that he is a
scooter driver. [Ans. 3
44]
8. A factory has three machines X, Y and Z producing 1000, 2000
and 3000 bolts per day respectively. The machine Xproduces 1%
defective bolts, Y produces 1.5% and Z produces 2% defective bolts.
at the end of a day, a bolt is drawnat random and is found
defective. What is the probability that this defective bolt has
been produced by the machine X?
[Ans. 0.1]9. A firm produces steel pipes in three plants A, B
and C with daily production of 500, 1000 and 2000 units
respectively. It
is known that fractions of defective output produced by the
three plants are respectively 0.005, 0.008 and 0.010. a pipeis
selected at random from a days total production and found to be
defective. What is the probability that it came fromfirst plant?
[Ans. 3/49]
10. Find the probability distribution of the number of tails
when five fair coins are tossed simultaneously.
[Ans.
0 1 2 3 4 51 5 10 10 5 1
32 32 32 32 32 32
]
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11. Find the probability distribution of the number of successes
in two tosses of a die when a success is defined as getting
a value 5 or 6. [Ans.
0 1 24 4 19 9 9
]
12. Find the probability distribution of the number of doublets
in four throws of a pair of dice.
[Ans.
0 1 2 3 4625 500 150 20 1
1296 1296 1296 1296 1296
]
13. Three cards are drawn successively with replacement from a
well-shuffled deck of 52 cards. A random variable Xdenotes the
number of spades in the three cards. Determine the probability
distribution and sketch its graph.
[Ans.
0 1 2 327 27 9 164 64 64 64
]
14. An urn contains 4 white and 3 red balls. Find the
probability distribution of the number of red balls when 3 balls
are
drawn one by one with replacement. [Ans.
0 1 2 364 144 108 27
343 343 343 343
]
15. A box contains 12 bulbs out of which 3 are defective . A
sample of 3 bulbs is selected from the box. Let X denote thenumber
of defective bulbs in the sample. Find the probability distribution
of X.
[Ans.
X 0 1 2 3 P(X) 84
220 108
220 27
220 1
220 ]
16. A fair coin is tossed 6 times. What are the parameters of
this binomial experiment. What is the probability of gettingexactly
4 heads? Atleast 4 heads. If the experiment is repeated 320 times,
how many times do you expect to get exactly
4 heads? Atleast 4 heads. [Ans. Parameters are n = 6, p = 12
. Thus, the binomial distribution is 61 1
2 2 Or X ~ B
16,2
;
P(X = 4) = 1564 and P(X 4) =
1132 . When the experiment is repeated 320 times, we expect
heads to come exactly 4 times
in 75 cases and atleast 4 times in 110 cases.]17. If X denotes
the number of heads in a single toss of 4 fair coins, find (i) P(X
= 3) (ii) P(X < 2) (iii) P (X 2)
(iv) P (1 < X 3). [Ans. (i) 14
(ii) 5
16 (iii) 1116 (iv)
58 ]
18. A pair of dice is thrown 7 times. If getting a total of 7 is
considered a success, what is the probability of (i) no sucess(ii)
6 successes (iii) atleast 6 sucesses (iv) at most 6 sucesses.
[Ans. (i) 75
6 (ii)
71356 ] (iii)
516 (iv)
7116 ]
19. In a box containing 100 bulbs, 10 are defective. What is the
probability that out of a sample of 5 bulbs (i) none is
defective (ii) exactly 2 are defective? [Ans. (i) 59
10 , (ii) (0.00729)]
20. There are 6% defective items in a large bulk of items. What
is the probability that a sample of 8 items will not includemore
than 1 defective item? [Ans. 71.42 (0.94) ]
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21. Five cards are drawn successively with replacement from a
well-shuffled deck of 52 cards. What is the probability that(i) all
the five card are spades (ii) only 3 cards are spades (iii) none is
spade ?
[Ans. (i) 51
4 , (ii)
51904 , (iii)
534 ]
22. Find the probability of guessing correctly atleast 6 of the
10 answers on a true-false examination.
[Ans. 193512 ]
23. A box contains 100 tickets, each bearing a different number
from 1 to 100. If 5 tickets are drawn successively withreplacement
from the box, find the probability that all tickets bear numbers
divisible by 10.
[Ans. 51
10 ]
24. The probability of a man hitting a target is 14
. If he fires 7 times, what is the probability of his hitting
atleast twice?
[Ans. 81818192 ]
25. The probability of a man hitting a target is 14
. Find the minimum no. of trials so that the probability of his
hitting the
target atleast once is greater than 23 ? [Ans. 4]
26. If the mean and variance of a Binomial distribution are
respectively 9 and 6, find the distribution.
[Ans. n = 27, p = 13 , q =
23 ]
27. The mean of a Binomial distribution is 40 and standard
deviation is 6. Calculate n, p, q.
[Ans. 1 9400, ,
10 10n p q ]
28. A die is thrown 20 times. Getting a number greater than 4 is
considered a success. Find the mean and the variance of the
number of successes. [Ans. 220 406.67, 4.443 9
]29. A die was thrown 800 times and 5 or 6 was obtained 220
times. Is the die biased?
[Ans. Yes, definitely baised, as the difference between observed
value and mean is 46 .]
30. If the sum of the mean and the variance of a Binomial
distribution for 54 trials is 30, find the distribution.
[Ans. n = 54, p = 13 , q =
542 2 1;3 3 3 ]