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Department of Mathematics, IIT Madras

MA2040 - Probability, Statistics and Stochastic Processes

January - May 2016Solutions to Problem Set I

1. ={1, 2, 3, 4, 5, 6}. Let us denote P({i}) = P(i).LetP(1) =P(3) =P(5) =p.This implies thatP(2) =P(4) =P(6) = 2p.By Normalization axiom 3p+ 6p= 1.Thereforep = 19 .Thus P(1) =P(3) =P(5) = 19 andP(2) =P(4) =P(6) =

29

.

P(outcome less than 4) = P({1, 2, 3})

=P(1) +P(2) +P(3)

=

4

9 .

2(a). S1, S2, , Sn are such that Si Sj = for any i, j(i=j) andS1 S2 Sn= .Now, A = (S1 A) (S2 A) (Sn A) (See Figure (i) below).P(A) =

ni=1P(Si A) (by finite additivity).

Figure (i) Figure (ii) Figure (iii)

2(b). Note that = (B CC) (BC C) (B C) (BC CC) (See Figure(ii) above). By 2(a) we have,

P(A) = P(ABCC)+P(ABCC)+P(ABCCC)+P(ABC) (1)

Note that(AB CC) (AB C) = (AB) (CCC) = (AB) =AB.

Further (ABCC) and (ABC) are disjoint (See Figure (iii) above).SimilarlyA C= (A BC C) (A B C). Thus

P(A B CC) = P(A B) P(A B C) (2)P(A BC C) = P(A C) P(A B C). (3)

Using (1),(2) and (3) we get

P(A) = P(A B) +P(A C) +P(A BC CC) P(A B C).

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3(a). We know thatP(A B) = P(A) +P(B) P(A B).From this it follows that

P(A B) = P(A) +P(B) P(A B)

P(A) +P(B) 1 (sinceP(A B) 1).

3(b). We prove by induction.We know by 3(a) that

P(A1 A2) P(A1) +P(A2) 1.

This shows that 3(b) is true for n = 2. Assume that 3(b) is true for (n1)events. That is,

P(A1 A2 An1) P(A1) +P(A2) + +P(An1) (n 2).

Now consider the case ofn events.

P((A1 A2 An1) An)

P(A1 A2 An1) +P(An) 1 (By 3(a))

P(A1) +P(A2) + +P(An1) (n 2) +P(An) 1 (By hypothesis)

=P(A1) +P(A2) + +P(An1) +P(An) (n 1).

4. Hint: Use induction.

5(a). A=the event that roll results in a sum of 4 or less.={(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}

B= the event that doubles are rolled.={(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}A B = {(1, 1), (2, 2)}Since all elementary outcomes are equally likely, P(A B ) = 236 andP(A) = 636 .

By definition of conditional probabilityP(B| A) = P(AB)P(A) = 2/366/36 =

13 .

5(b). A=the event that two dice land on different numbers.This mean P(A) = 30/36.B= the event that one die roll is a 6={(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}.A B = {(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}

P(A B) = 10/36.P(B| A) = 10/3630/36 =

13 .

6. Assume that the batch contains exactly five defective items.Four randomly selected items contains zero defective in

954

50

=954

ways.

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From 100 items any four items can be selected in

1004

ways.

The probability that none of the item is defective =

(954 )

(1004 ) = 95949392

100999897 0.812.

(Alternative Way)Define eventsA1= the first item chosen randomly is not defective.A2= second item is not defective.A3= third item is not defective.A4= fourth item is not defective.Now by multiplication ruleP(A1 A2 A3 A4)= Probability that all four items are not defective.= P(A1).P(A2 | A1).P(A3 | A1 A2).P(A4 | A1 A2 A3)= 95100 .

9499

.9398

.9297 0.812.

7. Probability that out ofn1 voice users , k1 want to connect to the systemisn1k1

pk11 (1 p1)

n1k1 , k1 = 0, 1, 2,...n1.In a similar way the probability that out ofn2 data users k2 (data users)want to connect to the system isn2k2

pk22 (1 p2)

n2k2 , k2 = 0, 1, 2,...n2.Since above two events are independent the probability thatk1 voice usersandk2 data users simultaneously want to connect to that system isn1k1

pk11 (1 p1)

n1k1 .n2k2

pk22 (1 p2)

n2k2 .

The total capacity used by k1 voice users and k2 data users isk1r1 + k2r2.Probability that requirements exceeds capacity is

{all k1, k2 such thatk1r1+k2r2 > c}

[

n1k1

pk11 (1p1)

n1k1 ][

n2k2

pk22 (1p2)

n2k2 ]

8. As per the question, qn is the probability of obtaining even number ofheads inn independent tosses.qn =P(the n

th toss is a head and there are odd number of heads in first(n1) tosses)+P(thenth toss is a tail and there are even number of headsin first (n 1) tosses )Since all the tosses are independent, we haveqn=p.P(there are odd number of heads in first (n 1) tosses)+(1 p).P( there are even number of heads in first (n 1) tosses )

=p.(1 qn1) + (1 p).qn1 = p+ (1 2p)qn1.From this recursion, we get the required formula for qn using the fact thatq0 = 1.

9. Let Ebe the event of getting 6 in a roll of a pair of dice

E= {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

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LetFbe the event of getting 7 in a roll of a pair of dice

F ={(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

Since all outcomes are equally likely,

P(E) = 5

36 and P(F) =

6

36=

1

6

A wins if he gets 6 before B gets a 7. Since both players roll the diealternately, A can win on 1st, 3rd, 5th rolls.Probability of winning ofA = A wins on first trial + A failsB failsAwins + A failsB fails A fails B fails A wins + cdots

= 5

36+

31

36

5

6

5

36+

31

36

5

6

31

36

5

6

5

36+

= 5

36

1 +

31

36

5

6+

31

36

5

6

2+

31

36

5

6

3+

= 5

36

1

1 3136

56

=5 6 3636 61

=30

61

10. Let A be the event that an increase in property tax is opposed.LetEbe the event that a registered voter is a property ownerThenEc = the event that a registered voter is not a property ownerNow it is given that P(A|E) = 0.6.This also implies thatP(Ac|E) = 0.4.It is also given that P(Ac|Ec) = 0.8, P(E) = 0.65. We need to find the

probability that an increase in property tax is favored i.e; P(Ac

)

P(Ac) = P(Ac E) +P(Ac Ec)

=P(Ac|E)P(E) +P(Ac|Ec)P(Ec)

= 0.4 0.65 + 0.8 0.35

= 0.54.

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