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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

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Session
Session

Probability - 3


Session objectives
Session Objectives

  • Random Variable

  • Probability Distribution

  • Applications of Probability

  • Class Exercise


Random variable
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
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

P(X = 0) = Probability of getting no head = P(TTT) =

P(X = 1) = Probability of getting one head

= P(HTT or THT or TTH ) =

P(X = 2) = Probability of getting two heads

= P(HHT or THH or HTH ) =

P(X = 3) = Probability of getting three heads

= P(HHH) =

Probability Distribution

Now, the probability of each outcome in the example is


Mathematics
Cont.

Probability distribution of X is:

P(X) is called the probability distribution of the random variable X.


Example 1
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
Solution (Cont.)

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 cont1
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)

When X = 2

Both the cards are aces

Hence, the probability distribution of X is


Example 2
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 cont2
Solution (Cont.)

P(X = 0) = Probability of getting no bad orange

= Probability of getting 2 good oranges

P(X = 1) = Probability of getting one bad orange


Solution cont3
Solution (Cont.)

P(X = 2) = Probability of getting two bad oranges

The probability distribution of X is given by


Example 3

Probability that the number appearing on the top is less than 3 in a

single through of a die

= Probability of success

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.


Solution cont4

Probability of not getting a success than 3 in a

Solution (Cont.)


Solution cont5
Solution (Cont.) than 3 in a

The probability distribution of X is given by


Example 4
Example –4 than 3 in a

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.


Solution cont6
Solution (Cont.) than 3 in a


Solution cont7
Solution (Cont.) than 3 in a

The probability distribution of X is given by


Example 5
Example –5 than 3 in a

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.


Solution cont8
Solution (cont.) than 3 in a


Solution cont9
Solution (cont.) than 3 in a

The probability distribution of X is given by


Applications of probability
Applications of Probability than 3 in a

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
Example –6 than 3 in a


Solution
Solution than 3 in a

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 cont10
Solution (Cont.) than 3 in a

P(circuit in working condition)

= P(S1, S2 and S3 working together)

[As working of one switch does not effect the working of the other i.e.

independent events]


Example 7
Example –7 than 3 in a

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 cont11
Solution (Cont.) than 3 in a

By addition theorem


Example 8
Example-8 than 3 in a

  • 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 cont12
Solution (Cont.) than 3 in a

P(A) = P(B) = 0.95 [Given]

(i) P(neither of these is available when needed)

[As availability of one engine is independent of the availability

of the other]

= (0.05) × (0.05) = 0.0025


Solution cont13
Solution (Cont.) than 3 in a

(ii) P(exactly one engine is available when needed)


Example 9

A company has estimated that the probabilities of success for

three products introduced in the market are

respectively. Assuming independence, find

(i) the probability that the three products are successful.

(ii) the probability that none of the products is successful.

Example –9

Solution: Let event A = First product is successful

event B = Second product is successful

event C = Third product is successful


Solution cont14
Solution (Cont.) for

(i) P(three products are successful)


Solution cont15
Solution (Cont.) for

(ii) P(none of the products are successful)


Example 10
Example –10 for

  • 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

  • the couple will be alive 25 years hence,

  • none of them will be alive 25 years hence,

  • (iii) at least one of them will be alive 25 years hence.


Solution1
Solution for

Let event A = the husband will be alive 25 years hence

event B = the wife will be alive 25 years hence

(i) P(couple will be alive 25 years hence)


Solution cont16
Solution (cont.) for

(ii) P(none of them will be alive 25 years hence)


Solution cont17
Solution (cont.) for

(iii) P(at least one of them will be alive 25 years hence)