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**Q. OTHER DEFINITIONS OF PROBABILITY**

**(a)Axiomatic probability :** Axiomatic approach is another way of describing probability of an event, in this approach some axioms or rules are depicted to assign probabilities.

Let S be the sample space of a random experiment. The probability P is a real valued function whose domain is the power set of S and range is the interval [0, 1] satisfying the following axioms:

(i)For any event E, P (E) ≥ 0 (ii) P(S) = 1

(iii)If E and F are mutually exclusive events, the P (E U F) = P(E) + P(F)

It follows from (iii) that P(φ) = 0

Let S be a sample space containing outcomes w_{1}, w_{2}..........., w_{n} i.e, S = {w_{1}, w_{2}, ............w_{n}}

It follows from the axiomatic definition of probability that :

(i)0 ≤ P(w_{i}) ≤ 1 for each w_{i} ∈ S

(ii)P(w_{1}) + P(w_{2}) + ..........+P(w_{n}) = 1

(iii)For any event A, P(A) = ∑ P(w_{i}), w_{i} ∈ A

**(b)Empirical probability :** A method which can be adopted in the example given above is to throw the dart several times (each throw is a trial) and count the number of times you hit the bull's-eye (a success) and the number of times you miss (a failure). Then an empirical value of the probability that you hit the bull's - eye with any one throw is

If the number of throws is small this does not give a particular good estimate but for a large number of throws the result is more reliable.

When the probability of the occurrence of an event A cannot be worked out exactly, an empirical value can be found by adopting the approach described above, that is :

(i) making a large number of trials (i.e. set up an experiment in which the event may, or may not, occur and note the outcome)

(ii) counting the number of times the event does occur, i.e. the number of successes,

(iii) calculating the value of

The probability of then event A occurring is defined as

mean that the number of trials is large (but what should be taken as ‘large’ depends on the problem).

**R. IMPORTANT POINTS**

(a)

(b) are mutually exclusive events then

(c) Let A & B are two events corresponding to sample space S then P(S|A) = P(A|A) = 1

(d) Let A and B are two events corresponding to sample space S and F is any other event s.t.

(e)

(f)

**Ex.27 A, B, C in order cut a pack of cards, replacing them after each cut, on the condition that the first who cuts a spade shall win a prize; find their respective chances. Sol.**Let p be the chance of cutting a spade and q be the chance of not cutting a spade from a pack of 52 cards. Then

Now A will win a prize if he cuts spade at 1st, 4th, 7th, 10th turns, etc. Note that A will get a second chance if A, B, C all fail to cut a spade once and then A cuts a spade at the 4th turn.

Similarly he will cut a spade at the 7th turn when A, B, C fail to cut spade twice, etc.

Hence A’s chance of winning the prize =

Similarly B’s chance =

and C’s chance =3/4 of B’s chance =

**Ex.28 (a) If p and q are chosen randomly from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, with replacement, determine the probability that the roots of the equation x ^{2} + px + q = 0 are real. (b) Each coefficient in the equation ax^{2 }+ bx + c = 0 is determined by throwing at ordinary die. Find the probability that the equation will have equal roots. Sol.**(a) If roots of x

Both p, q belongs to set when p = 1, no value of q from S will satisfy (i)

p = 2q = 1 will satisfy1 value

p = 3q = 1, 22 value

p = 4q = 1, 2, 3, 44 value

p = 5q = 1, 2, 3, 4, 5, 66 value

p = 6q = 1, 2, 3, 4, 5, 6, 7, 8, 9, 9 value

For p = 7, 8, 9, 10 all the ten values of q will satisfy.

Sum of these selections in 1 + 2 + 4 + 6 + 9 + 10 + 10 + 10 = 62

But the total number of selections of p and q without any order is 10 × 10 = 100

Hence the required probability is 62/100 = 0,62

(b) Roots equal ⇒ b_{2} - 4ac = 0 ....(i)

Each coefficient is an integer, so we consider the following cases :

b = 1No integral values of a and c

b = 21 = ac∴ 1/4 (1, 1)

b = 39/2 = acNo integral values of a and c

b = 44 = ac∴ (1, 4), (2, 2), (4, 1)

b = 525/2 = acNo integral values of a and c

b = 69 = ac∴ (3, 3)

Thus we have 5 favourable way for b = 2, 4, 6,

Total number of equations is 6.6.6 = 216∴ Required probability is \(5 \over 216\)

**Ex.29In a test an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is 1/3 and the probability that he copies the answer is \(1 \over 6\) . The probability that his answer is correct given that he copied it, is 1/8. Find the probability that he knew the answer to the question given that he correctly answered it.**

**Sol.**Let A_{1} be the event that the examinee guesses that answer ; A_{2} the event that he copies the answer and A_{3} the event that he knows the answer. Also let A be the event that he answers correctly. Then as given, we have

[We have assumed here that the events A_{1}, A_{2} and A_{3} are mutually exclusive and totally exhaustive.]

Now (as given)

Again it is reasonable to take the probability of answering correctly given that he knows the answer as 1, that is P( A | A _{3} ) = 1 . We have to find P( A_{ 3} | A ) .

By Baye’s theorem, we have

**Ex.30 A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events A, B, C are defined as A = {The first bulb is defective} B = {The second bulb is non-defective} C = {The two bulbs are both defective or both non-defective} Determine whether (i) A, B, C are pairwise independent, (ii) A, B, C are independent.**

**Sol. **We have

A ∩ B is the event that first bulb is defective and second is non-defective.

A ∩ C is the event that first bulb is defective and second is non-defective.

Similarly P(B ∩ C) = 41 . Thus we have P( A ∩ B) = P( A ) . P(B) ; P( A ∩ C) = P( A ) . P(C) ; P(B ∩ C) = P(B) . P(C)

∴ A, B and C are pairwise independent. There is no element in A ∩ B ∩ C

∴ P(A ∩ B ∩ C) = 0

∴ P( A ∩ B ∩ C) ≠ P(A ) . P(B) . P(C)

Hence A, B and C are not mutually independent.

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