Transcript Document

Mathematics
Session
Probability - 3
Session Objectives
 Random Variable
 Probability Distribution
 Applications of Probability
 Class Exercise
Random Variable
Let us consider a random experiment of tossing three coins
(or a coin is tossing three times).
Then sample space is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
We are interested in the number of heads in each outcome. Let X
denote the number of heads in each outcome, then X takes the values
Random Variable (Cont.)
X (HHH) = 3, X(HHT) = 2, X(HTH) = 2, X(THH) = 2 , X(HTT) = 1,
X (THT) = 1, X (TTH) = 1 and X(TTT) = 0 i.e. X = 0, 1, 2 or 3
X is a variable obtained from the random experiment and called
random variable.
Probability Distribution
Now, the probability of each outcome in the example is
1
P(X = 0) = Probability of getting no head = P(TTT) =
8
P(X = 1) = Probability of getting one head
= P(HTT or THT or TTH ) =
3
8
P(X = 2) = Probability of getting two heads
3
= P(HHT or THH or HTH ) =
8
P(X = 3) = Probability of getting three heads
1
= P(HHH) =
8
Cont.
Probability distribution of X is:
X
0
1
P(X)
8
1
3
8
2
3
8
3
1
8
P(X) is called the probability distribution of the random variable X.
Example –1
Two cards are drawn successfully with replacement from a well shuffled
pack of 52 cards. Find the probability distribution of the number of aces.
Solution: The total number of cards in a pack of cards is 52.
Let X denote the number of aces, then X can take the values 0, 1 or 2.
When X = 0
Both the cards are non-aces.
P(X = 0) = P(First drawn card is non-ace and second drawn card
is also non-ace)
Solution (Cont.)
= P(First drawn card is non- ace) ×P(Second drawn card is non-ace)
=

48 48
×
52 52
[As the cards are drawn with replacement]
12 12 144


13 13 169
When X = 1
One card is an ace and other is a non-ace
P(X = 1) = P(First drawn card is ace and second drawn card is
non-ace or first drawn card is non-ace and second
drawn card is ace)
Solution (cont.)
= P(First drawn card is ace and second drawn card is non-ace)
+ P(first drawn card is non-ace and second drawn card is ace)
=
4 48
24
×
×2 =
52 52
169
When X = 2
Both the cards are aces
P  X =2 =
4
4
1
×
=
52 52 169
Hence, the probability distribution of X is
X
0
1
P(X)
144/169 24/169
2
1/169
Example –2
Four bad oranges are mixed accidently with 16 good oranges.
Find the probability distribution of the number of bad oranges in
a draw of two oranges.
Solution:
Total number of oranges = 4 (bad oranges) +16 (good oranges) = 20
Let X denote the number of bad oranges, then X can take the
values 0, 1 or 2.
Solution (Cont.)
P(X = 0) = Probability of getting no bad orange
= Probability of getting 2 good oranges
=
16
C2 12
=
20
C2 19
P(X = 1) = Probability of getting one bad orange
4
C1 × 16 C1 32
=
=
20
C2
95
Solution (Cont.)
P(X = 2) = Probability of getting two bad oranges
=
4
C2
3
=
20
C2 95
The probability distribution of X is given by
X
0
1
2
P(X)
12/19
32/95
3/95
Example –3
A fair die is tossed twice. If the number appearing on the top is less
than 3. It is a success. Find the probability distribution of number of
successes.
(CBSE 2004)
Solution:
Let X denote the number of successes (getting a number less 3) when
a fair die is tossed twice. Then X takes the values 0, 1 or 2.
Probability that the number appearing on the top is less than 3 in a
single through of a die
= Probability of success =
2 1
=
6 3
Solution (Cont.)
1
3
Probability of not getting a success =1- =
2
3
2 2 4
P  X =0 =P no success = × =
3 3 9
P  X =1 =P one success and other no success
1 2 2 1 4
= × + × =
3 3 3 3 9
1 1 1
P  X =2 =P both successes = × =
3 3 9
Solution (Cont.)
The probability distribution of X is given by
X
0
1
2
P(X)
4/9
4/9
1/9
Example –4
From a lot of 30 bulbs which include 6 defectives, a sample of
4 bulbs is drawn at random with replacement. Find the probability
distribution of the number of defective bulbs.
(CBSE 2004)
Solution:
Total number of bulbs = 30 = 6 defective bulbs + 24 non-defective bulbs
Let X denote the number of defective bulbs. Then X can take the
values 0, 1, 2, 3 or 4.
P  a defective bulb  =
6
1
=
30 5
1 4
P  a non- defective bulb =1- 
5 5
Solution (Cont.)
4 4 4 4 256
P  X =0 =P no defective bulb  = × × × =
5 5 5 5 625
P  X =1 =P 1 defective bulb and 3 non- defective bulbs
 1 4 4 4  256
= 4 × × ×  =
 5 5 5 5  625
P  X =2 =P 2 defective bulbs and 2 non- defective bulbs
 1 1 4 4  96
=6  × × ×  =
 5 5 5 5  625
Solution (Cont.)
P  X =3 =P 3 defective bulbs and 1 non- defective bulb
 1 1 1 4  16
= 4 × × ×  =
 5 5 5 5  625
P  X = 4 =P  4 defective bulbs
1 1 1 1
1
= × × × =
5 5 5 5 625
The probability distribution of X is given by
X
0
1
2
3
4
P(X)
256/ 625
256/ 625
96/625
16/625
1/625
Example –5
An urn contains 4 red and 3 blue balls. Find the probability
distribution of the number of blue balls in a random draw of
3 balls with replacement.
Solution:
Total number of balls = 4 red + 3 blue = 7 balls
Let X denote the number of blue balls. Then X can take the
values 0, 1, 2 or 3.
P  a blue ball =
3
4
and P a red ball =
7
7
Solution (cont.)
P  X =0 =P no blue ball =P 3 red balls
4 4 4
64
= × × =
7 7 7 343
P  X =1 =P 1 blue ball and 2 red balls
 3 4 4  144
=3  × ×  =
 7 7 7  343
P  X =2 =P 2 blue balls and 1 red ball
 3 3 4  108
=3 × ×  =
 7 7 7  343
Solution (cont.)
P  X =3 =P 3 blue balls
3 3 3 27
= × × =
7 7 7 343
The probability distribution of X is given by
X
0
1
2
3
P(X)
64/343
144/343
108/343
27/343
Applications of Probability
Probability theory has diverse use in science and technology these
days. In Biology, genetics involves a lot of calculations using
probability. It is also used in various fields of engineering.
Example –6
What is the probability that a series circuit with
three switches S1, S2 and S3 with probabilities
1 1
3
, and respectively, of functioning will work?
3 2
4
S1
A
S2
S3
B
Solution
The circuit would work if all the switches S1, S2 and S3 work together.
Consider event S1 : switch S1 is working
event S2 : switch S2 is working
event S3 : switch S3 is working
Solution (Cont.)
P(circuit in working condition)
= P(S1, S2 and S3 working together)
=P S1  S2  S3 
=P S1 ×P S2  ×P S3 
[As working of one switch does not effect the working of the other i.e.
independent events]
1 1 3 1
= × × =
3 2 4 8
Example –7
The probability that a person visiting a dentist will have his teeth
cleaned is 0.44, the probability that he will have a cavity filled is
0.24. The probability that he will have his teeth cleaned or a cavity
filled is 0.6. What is the probability that a person visiting his dentist
will have his teeth cleaned and a cavity filled?
Solution: Let event A = the person will have his teeth cleaned.
event B = he will have the cavity filled.
Solution (Cont.)
P  A  =0.44, P B =0.24 and P  A  B =0.60
P(teeth cleaned and a cavity filled)=P  A  B
By addition theorem
P  A  B =P  A  +P B -P  A  B
 0.60=0.44+0.24-P  A  B
 P  A  B =0.08
Example-8
A town has two fire extinguishing engines functioning independently.
The probability of availability of each engine, when needed is 0.95.
What is the probability that
(i)
neither of these is available when needed.
(ii) exactly one engine is available when needed.
Solution: Let event A = availability of one engine.
event B = availability of other engine.
Solution (Cont.)
P(A) = P(B) = 0.95
[Given]
 P  A' =1-P  A  and P B' =1-P B
 P  A' =0.05 and P B' =0.05
(i)
P(neither of these is available when needed)
=P  A'  B'
=P  A' ×P B'
[As availability of one engine is independent of the availability
of the other]
= (0.05) × (0.05) = 0.0025
Solution (Cont.)
(ii)
P(exactly one engine is available when needed)
=P  A  B -P  A  B
=P  A  +P B -2P  A  B
=P  A  +P B -2P  A ×P B
=0.95+0.95-2 0.95×0.95
= 0.095

A, B are independent events
Example –9
A company has estimated that the probabilities of success for
three products introduced in the market are
1 2
2
, and
3 5
3
respectively. Assuming independence, find
(i) the probability that the three products are successful.
(ii) the probability that none of the products is successful.
Solution: Let event A = First product is successful
event B = Second product is successful
event C = Third product is successful
Solution (Cont.)
1
2
2
P  A  = , P B = and P C =
3
5
3
(i) P(three products are successful)
=P  A  B  C 
=P  A  ×P B×P C 
1 2 2
4
= × × =
3 5 3 45

A, B, C are independent events
Solution (Cont.)
(ii) P(none of the products are successful)
=P  A'  B'  C'
=P  A' ×P B'×P C'
 1  2  2
= 1-  ×1-  ×1- 
 3  5  3
2 3 1 2
= × × =
3 5 3 15

A, B, C are independent events
Example –10
The odds against a husband who is 45 years old, living till he is 70
are 7 : 5 and the odds against his wife who is now 36, living till she
is 61 are 5 : 3. Find the probability that
(i) the couple will be alive 25 years hence,
(ii) none of them will be alive 25 years hence,
(iii) at least one of them will be alive 25 years hence.
Solution
Let event A = the husband will be alive 25 years hence
event B = the wife will be alive 25 years hence
P A =
5
5
3
3
=
and P B =
=
7+5 12
5+3 8
(i) P(couple will be alive 25 years hence)
=P  A  B 
=P  A  ×P B
=
5 3
5
× =
12 8 32

A, B are independent events
Solution (cont.)
(ii) P(none of them will be alive 25 years hence)
=P  A'  B'
=P  A' ×P B'
5   3

= 1× 1- 

 12   8 
=
7 5 35
× =
12 8 96

A, B are independent events
Solution (cont.)
(iii) P(at least one of them will be alive 25 years hence)
=1- P  A'  B'
=1-P  A'×P B'
5   3

=1- 1×1- 

 12   8 
=1-
35 61
=
96 96

A, B are independent events
Thank you