CHAPTER - 3 Probability

Size: px

Start display at page:

Download "CHAPTER - 3 Probability"

Simon Carter
5 years ago
Views:

1 CHAPTER - 3 Probability 3.10 Glossary of probability Terms A B: An event which represents the happing of at least one of the events A and B. A B: An event which represents the simultaneous happening of both the events A and B : A does not happen. : Neither A nor B happens B: A does not happen but B happens ( B) (A ): Exactly one of the two events A and B happens Addition theorem of probability The probability of occurrence of at least one of the two events A and B is given by: P(A B)=P(A)+P(B)-P(A B). Note The probability of happing of any one of the two mutually disjoint events is equals to the sum of their individual probabilities Independent Events Events are said to be independent of each other if happing of any one of them is not affected by and does not affect the happening of any one of the others. If a and B are independent events so that the probability of occurrence or nonoccurrence of A is not affected by occurrence or non-occrruence of B,then we have P( )=P(A) and P( )=P(B) Conditional Probability Let A and B be two events such that P(A) >0. Denote by P the probability of B given A has occurred. Since A is known to have occurred, it becomes the new sample space replacing the original S. From this we are led to the definition P(B / A) =P(A B)/P(A) K L University, Vaddeswaram Page 51

2 (OR) P(A B) =P(A) P(B/ A) that the probability that both A and B occur is equal to the probability that A occurs times the probability that B occurs given that A has occurred. We call P (B/ A) the conditional probability of B given A, i.e., the probability that B will occur given that A has occurred Multiplication theorem of probability The probability of simultaneous happing of two events A and B is given by : P(A B)=P(A)P( ) ; p(a) 0 P(A B)=p(B)p( );p(b) 0 Where P( ) is the conditional probability of happening of B under the condition that A has happened and P( ) is the condition probability of happening of A under the condition that B has happened Generalisation Of Multiplication Theorem Of Probability For three events A 1, A 2,A 3 we have P(A B C)=P(A 1 )P(A 2 / A 1 )P(A 3/ A 1 A 2 ) 3.16 Multiplication theorem on Independent events Two events A and B are independent P(A B)=P(A).P(B) Properties 1. P( )=1 P(A) 2. P( )=P(B) P(A B) 3. P( )=P(A) P(A B) 4. If A is subset of B then P(A) P(B). Example 3.24 A card is drawn from a well shuffled pack of playing cards. Find the probability that it is either a diamond or a king. K L University, Vaddeswaram Page 52

3 Let A denote the event of drawing a diamond and B denote the event of drawing a king from a pack of cards. Then we have P(A)= and P(B)= and we want P(A B). P(A B) = P(A)+P(B)-P(A B)= + - P(A B) There is only one case favourable to the event A B viz.,king of diamond.hence, P(A B)= Substituting in above, we get P(A B)= + - = = = Example 3.25 If P(A)=0.4,P(B)=0.7 and p(at least one of A and B)=0.8, find P(only one of A and B). We are given:p(a)=0.4,p(b)=0.7 and P(A B)=0.8. The event only one of A and B, can materialize in the following mutually disjoint ways: (i)a and not B i.e., A (ii) Not A and B i.e., B. Hence the required probability is given by: P=P[only one of A and B] =P(i)+P(ii) =P B) +P(A =P (B)- P(A B)+P(A)-P(A B) = P(A B) We have P(A B) = P(A)+P(B)-P(A B) P(A B)= P(A)+P(B)- P(A B)= =0.3 Substituting in above, we get: P= (0.3) =0.5 Example 3.26 Let A and B be two possible outcomes of an experiment and suppose P(A)=0.4, P(A B)=0.7 and P(B)=p (i)for what choice of p are A and B mutually exclusive? K L University, Vaddeswaram Page 53

4 (ii) For what choice of p are A and B independent? (i) We have P(A B)= P(A)+P(B)-P(A B) P(A B)= P(A)+P(B)- P(A B)=0.4+P-0.7=P-0.3 If A and B are mutually exclusive,then P(A B)=0 => P-0.3=0 => P=0.3 (ii)a and B are independent if and only if P(A B)=P(A).P(B) P-0.3=(0.4) P =>(1-0.4)P=0.3 =>( 0.6 )P=0.3 => p= =0.5. Example 3.27 If A&B are mutually exclusive events P(A)=0.29, P(B)=0.43 find 1) P( ) 2) P(AUB) 3) P(A ) 4) P( ) 1) P(A) =1-0.29=0.71 2) P(AUB)=P(A)+P(B)= =0.72 3) P(A )=P(A)-P(A B) =0.29 4) P( )= P( ) =1-P(AUB) = =0.28 Example 3.28 The probability that a new airport will get an award for its design is 0.16.The probability that it will an award for the efficient use of materials is 0.24 and the probability that it will get both awards is 0.11 a) What is the probability that it will get atleast one of the two awards? b) What is the probability that it will get only one of the awards? P(D)=0.16, P(M)=0.24, P(D M)=0.11 a) P(DUM)=P(D)+P(M)-P(D M)= =0.29 b) P(DUM)-P(D M)= =0.18 K L University, Vaddeswaram Page 54

5 Example 3.29 A fair die is tossed twice. Find the probability of getting a 4, 5, or 6 on the first toss and a 1, 2, 3, or 4 on the second toss. Let A 1 be the event 4, 5, or 6 on first toss, and A 2 be the event 1, 2, 3, or 4 on second toss. Then we are looking for P(A 1 A 2 ). Each of the 6 ways in which a die can fall on the first toss can be associated with each of the 6 ways in which it can fall on the second toss, a total of 6. 6 = 36 ways, all equally likely. Each of the 3 ways in which A 1 can occur can be associated with each of the 4 ways in which A2 can occur to give 3. 4 = 12 ways in which both A 1 and A 2 can occur. Then P(A 1 A 2 )=12/36 =1/3. This shows directly that A1 and A2 are independent since P(A 1 A 2 )=1/3=(3 /6) (4/6)= P(A 1 )P(A 2 ) Example 3.30 If P(C) =0.65, P(D)=0.40, P(C D)=0.24. Are the events C& D independent P(C D)= P(C) P (D)= =0.26 clearly, P(C D) P(C) P(D) i.e C,D are not independent Example 3.31 P(A)=0.5, P(B)=0.3,P(A B)=0.15 verify that a. P(A/B) =P(A) b. P(A/ )=P(A) c. P(B/A)=P(B) d. P(B/ )=P(B) a) P(A/B) =P(A B)/P(B) =0.15/0.3 =0.5 K L University, Vaddeswaram Page 55

6 There fore P(A/B) =P(A) b) P(A/ ) =P(A )/P(B) =(P(A)-P(A B))/(1-P(B)) =( )/(1-0.3) =0.5 c) P(B/A) =P(B A)/P(A) =0.15/0.5=0.3 There fore P(B/A)=P(B) d) P(B/ ) =P(B )/P(A) =(P(B)-P(B A))/(1-P(A)) =( )/(1-0.5) =0.3 There fore P(B/ )=P(B) Example 3.32 If the probabilities are 0.58,0.25,&0.19 that a person in a certain income will invest money market funds, common stocks or both find the probabilities that a person in that income a)who invest in money market funds will also invest in common stocks b) Who invest common stock also invest in money market funds. Given that P(MM)=0.58 P(CS)=0.25 P(MM CS)=0.19 a) Probability of investing money in money marketing funds will also invest in common stocks. P(CS/MM) =P(CS MM)/P(MM) =0.19/0.58 =0.327 b). Probability of investing in common stock will also invest in MM funds. P(MM/CS)=P(MM CS)/P(CS) =0.19/0.25 =0.76 K L University, Vaddeswaram Page 56

7 Example 3.33 The probability that a regularly scheduled flight depends on time is P(D)=0.83.The probability that it arrives on time P(A)=0.82 & The probability that it departs & arrives on time is P(D A)=0.78.Find the probability that a plane. a) Arrives on time given that it departed on time. b) Departed on time given that it has arrived on time. c) Arrives on time given that it did not depart on time. P(D)=0.83 the probability that the flight depends on time a) The probability that a plane arrives on time given that it departed on time is P(A/D)=P(A D)/P(D) =0.73/0.83 =0.93 b)probability that a plane departed on time given that it arrives on time. P(D/A)=P(D A)/P(A) =0.78/0.83 =0.95 c) P(A/D) =P(A )/P( ) =(P(A)-P(A D))/(1-P(D)) = / =0.23 Example 3.34 If the odds are 5 to 3 that an event M will occur and,2 to 1 that event N will occur and 4 to 1 that they will not both occur.are two events M&N independent? By the definition,p(m N)=P(M) P(N) P(M)=3/5=0.6 P(N)=1/2=0.5 P( )=P( ) ¼ = 1-P(MUN) ¼ = 1-{P(M)+P(N)-P(M N)} ¼ = 1-{ P(M N)} P(M N)= =0.75 K L University, Vaddeswaram Page 57

8 P(M N)=( )-0.75 =0.15 P(M N) =0.15 P(M).P(N) = =0.2 Therefore P(M N) P(M). P(N) M and N are not independent. Example 3.35 Two cards are drawn from a well-shuffled ordinary deck of 52 cards. Find the probability that they are both aces if the first card is (a) replaced, (b) not replaced. Let A 1 =event ace on first draw A 2 = event ace on second draw. Then we are looking for P(A 1 A 2 ) =P(A 1 ) P( ) (a) Since for the first drawing there are 4 aces in 52 cards, P(A 1 ) = 4/52. Also, if the card is replaced for the second drawing, then P(A 2 / A 1 ) = 4/ 52, since there are also 4 aces out of 52 cards for the second drawing. Then P(A 1 A 2 ) = P(A 1 ) P( ) = ( 4/52)(4/52)=1/169. (b) As in part (a), P(A1) = 4/52. However, if an ace occurs on the first drawing, there will be only 3 aces left in the remaining 51 cards, so that P(A2/ A1) = 3/ 51. Then P(A 1 A 2 ) = P(A 1 ) P( ) =(4/52)(3/51)=1/221. Example 3.36 Find the probability of a 4 turning up at least once in two tosses of a fair die. Let A 1 = event 4 on first toss A 2 = event 4 on second toss. Then A 1 A 2 =event 4 on first toss or 4 on second toss or both = event at least one 4 turns up, Events A 1 and A 2 are not mutually exclusive, but they are independent. P(A 1 A 2 ) =P(A 1 ) + P(A 2 ) - P(A 1 A 2 ) = P(A 1 ) + P(A 2 ) - P(A 1 )P( A 2 ) = = K L University, Vaddeswaram Page 58

CPT Section D, Quantitative Aptitude, Chapter 13 Prof(Dr.) P.R.Vittal

CPT Section D, Quantitative Aptitude, Chapter 13 Prof(Dr.) P.R.Vittal Random experiment Sample space Events Null event (Impossible event) 1.Classical probability 2.Statistical definition Modern definition(axiometric