Chapter 2: Probability - Short Lecture and Answers

Chapter 2 PROBABILITY Rules of Probability 1. P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 2.P(A ∩ B) = P(A) + P(B) – P(A ∪ B) 3.P(𝐀̅) = P( Ac ) = 1 – P (A) 4. P(A ∩ 𝑩̅) = P(A - B) = P(A) - P(A ∩ B) 5. P(𝑨̅ ∩ 𝑩) = P(B – A) = P(B) - P(A ∩ B) 6. P(𝐀̅ ∩ 𝐁̅)= 1 - P(A ∪ B) 7. P(A U B U C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C ) – P(B ∩ C) + P(A ∩ B ∩ C) If A & B are independent - P(A ∩ B) = P(A) * P(B) -P(A U B)= P(A) + P(B) – P(A)*P(B) If A & B are mutually exclusive ( disjoint ) -P(A ∩ B) = 0 -P(A U B)= P(A) + P(B) conditional probability of A given B 𝑝(𝐴|𝐵 ) = 𝑃 (𝐴 ∩ 𝐵 ) 𝑝(𝐵 ) [1] An integer between 1 and 100 is selected at random, find the probability of getting a perfect square if a- all integers are equally likely to be selected. b- all integers between 1 and 50 are twice as likely to occur as the rest. Solution: a) all integers are equally likely to be selected S = { 1 , 2 , 3 , 4 ,……. , 100 } A = { 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100 } P(A) = n(A) n(S) = 10 100 = 0.1 b- all integers from 1 to 50 are twice as likely to occur as the rest A = { 1 , 4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100 } 2p p 2p*50+p*50=1 , 150p=1 , p= 1 150 P(A) = 7 * 2 150 +3* 1 150 = 0.11333 [2] Three students A, B and C are in a swimming race. A and B have the same probability of winning and each is twice as likely to win as C. a- What is the probability that A does not win? b- What is the probability that B or C wins? Solution: P(A) = P(B) = 2P(C) P(A)+P(B)+P(C) = 1 2P(C) + 2P(C) + P(C) = 1 , 5P(C) = 1 P(C) = 1/5 , P(A) = 2/5 , P(B) = 2/5 a- What is the probability that A doesn’t win? ̅) = 1 – P(A) = 1 – 2/5 = 3/5 P(A b- What is the probability that B or C wins? P(B ∪ C) = P(B) + P(C) – P(B ∩ C) = 2/5 + 1/5 – 0 = 3/5 [3] In a certain population of women 4% have had breast cancer, 20% are smokers and 3% are smokers and have had breast cancer. A woman is selected at random from the population. What is the probability that: a- she has had breast cancer or smokes? b- she has had breast cancer and she is not a smoker ? Solution: breast cancer→ C , smoker→S P(C) = 0.04 , P(S) = 0.2 , P(C ∩ S) = 0.03 a- she has had breast cancer or smokes? P(C ∪ S) = P(C) + P(S) – P(C ∩ S) = 0.04 + 0.2 - 0.03 = 0.21 b- she has had breast cancer and she is not a smoker ? b) P(C ∩ S̅) = P(C) - P(C ∩ S) = 0.04 – 0.03 = 0.01 [4] The probability that a student, selected at random from a certain college, passes mathematics is 0.8, the probability that passes English is 0.6 and the probability of passing at least one of them is 0.9. a- What is the probability that he failed both Math. and English ? b- What is the probability that he failed Math only? c- If the student passed Math., what is the probability that he also passes English ? d- If the student failed Math., what is the probability that he also failed English ? Solution: Math→ M , English→ E P(M)=0.8 , P(E)=0.6 , P(M U E)=0.9 a- What is the probability that he failed both Math. and English ? ̅ ∩̅ P(M E) = 1 – P(M U E) = 1 – 0.9 = 0.1 b- What is the probability that he failed Math only? ̅ ∩ E) = P(E) - P(M ∩ E) P(M = 0.6 – ( 0.8 + 0.6 – 0.9 ) = 0.1 c- If the student passed Math., what is the probability that he also passes English ? P(E / M) = P( E ∩ M) P(M) = 0.8+0.6−0.9 0.8 = 0.625 d- If the student failed Math., what is the probability that he also failed English ? ̅ ̅ ̅/M ̅ ) = P( E ∩ M) = 1−P( E U M ) P(E ̅ P(M) 1 − P(M) 1−0.9 1 − 0.8 = 0.5 [5] If the three event A, B and C are independent with P(A)=1/4 , P(B) = 1/3 and P(C) = 1/5, find P(A  B  C). Solution: P(A U B U C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C ) – P(B ∩ C) + P(A ∩ B ∩ C) 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 𝟒 𝟑 𝟓 𝟒 𝟑 𝟒 𝟓 𝟑 𝟓 𝟒 𝟑 𝟓 = + + - ( ) * ( ) - ( )*( ) - ( )*( ) + ( )*( )*( ) = 0.6 [6] In a high school graduating class of 100 students, 54 studied mathematics, 69 studied English and 35 studied both mathematics and English. If one of these students is selected at random, find the probability that: a- the student took mathematics or English, b- the student did not take either of these subjects, c- the student took English but not mathematics. Solution: a- P( E U M ) = P(E) + P(M) – P( E ∩ M ) = 0.69 + 0.54 – 0.35 = 0.88 ̅∩M ̅ ) = 1 – P( E U M ) = 1 – 0.88 = 0.12 b- P(E ̅ ) = P(E) - P( E ∩ M ) = 0.69 – 0.35 = 0.34 c- P(E ∩ M [7] Let A and B events with P(A) = 1/2 , P(B) = 1/3 And P(A U B) = 7/12 , Find a) P(A / B) ̅) b) P(A / B ̅ / B) d) P(A ̅/B ̅) c) P(A Solution: a- P(A / B) = ̅) = b- P(A / B P(A∩B) P(B) ̅) P(A∩B ̅) P(B = = P(A)+P(B)−P(A U B) P(B) P(A)−P(A ∩ B) 1−P(B) = = 1 1 − 2 4 1 1− 3 1 1 7 + − 2 3 12 1 3 = 3 8 = 0.74 ̅ / B) = c- P(A ̅/B ̅) = d- P(A ̅ ∩B) P(A P(B) ̅ ∩B ̅) P(A ̅) P(B = = P(B)−P(A ∩ B) = P(B) 1 − P(A U B) 1− P(B) = 1 1 − 3 4 1 3 7 12 1 1− 3 1− = = 0.25 5 8 [8] Given P(A) = 0.5 and P(AUB) = 0.8 , find P(B) if: a) A and B are mutually exclusive b) A and B are independent c) P(A|B)=0.4. d) A is a subset of B. Solution: a) P(A U B) = P(A) + P(B) 0.8 = 0.5 + P(B) P(B) = 0.3 b) P(A∩B) P(B) = 0.4 , 0.5+P(B)−0.8 P(B) P(A)+P(B)−P(AUB) P(B) = 0.4 = 0.4 , 0.5 + P(B) – 0.8 = 0.4P(B) 0.6P(B) = 0.3 , P(B) = 0.5 c) P(A ∩ B) = P(A) * P(B) P(A ∩ B) = P(A) + P(B) – P(A U B) P(A)*P(B) = P(A) + P(B) – P(A U B) 0.5P(B) = 0.5 + P(B) – 0.8 0.5P(B) = 0.3 , P(B) = 0.6 d) P(A U B) = P(B) A is a subset of B. P(B) = 0.8 B A [9] Fifty-two percent of the students at a certain college are females. Five percent of the students in this college are majoring in computer science. Two percent of the students are women majoring in computer science. If a student is selected at random, find the conditional probability that (a) This student is female, given that the student is majoring in computer science. (b) This student is majoring in computer science, given that the student is female. Solution: C →computer science Female→ F P(F)=0.52 , P(C)=0.05 , (F∩ 𝐶)=0.02 (a) This student is female, given that the student is majoring in computer science. P(F/C)= 𝑝(𝐹∩𝐶) 𝐹(𝐶) 0.02 = 0.05 =0.4 (b) This student is majoring in computer science, given that the student is female. P(C/F)= 𝑝(𝐹∩𝐶) 𝐹(𝐹) = 0.02 0.52 =0.038 [10] A large industrial firm uses 3 local motels to provide overnight accommodations for its clients. From past experience it is known that 20%: of the clients are assigned rooms at the Ramada Inn, 50% at the Sheraton, and 30% at the Hilton. If the plumbing is faulty in 5% of the rooms at the Ramada Inn, in 4% of the rooms at the Sheraton, and in 8% of the rooms at the Hilton, what, is the probability that (a) a client will be assigned a room with faulty plumbing? (b) a person with a room having faulty plumbing was assigned accommodations at the Hilton? Solution: Client stays in Ramada Inn=R , P(R)=0.2 Client stays in Sheraton=S P(S)=0.5 , Client stays in Hilton=H , P(H)=0.3 Let faulty plumbing in hotels detoted by F P(F/R)=0.05 P(F/S)=0.04 P(F/H)=0.08 (a) a client will be assigned a room with faulty plumbing? P(F)= P(F|R)*P(R)+ P(F|S)* P(S)+ P(F|H)* P(H) =0.05*0.2+0.04*0.5+0.08*0.3 =0.054 (b) a person with a room having faulty plumbing was assigned accommodations at the Hilton? P(H|F)= 𝑃(𝐻∩𝐹 ) 𝑝(𝐹 ) = P(F|H)∗ P(H) 𝑝(𝐹 ) = 0.08∗0.3 0.054 =0.44 [11] Suppose that the four inspectors at a film factory are supposed to stamp the expiration date on each package of film at the end of the assembly line. John, who stamps 20% of the packages, fails to stamp the expiration date once in every 200 packages: Tom, who stamps 60% of the packages, fails to stamp the expiration date once in every 100 packages; Jeff, who stamps 15% of the packages, fails to stamp the expiration date once in every 90 packages; and Pat, who stamps 5% of the packages, fails to stamp the expiration date once in every 200 packages. If a customer complains that her package of film does not show the expiration date, what is the probability that it was inspected by John? Solution: We assume that John=J , Tom=T , Jeff=J1 , Pat=P S= package does not show the expiration date P(T)=0.6 , P(J)=0.2 , P(J1)=0.15 , P(P)=0.05 Tom, who stamps 60% of the packages, fails to stamp the expiration date once in every 100 packages 1 P(S|T)= 100 , P(S|J)= 1 200 , P(S|J1)= 1 , 90 P(S|P)= 1 200 If a customer complains that her package of film does not show the expiration date, what is the probability that it was inspected by John P(S)= P(S|T)*P(T)+ P(S|J)*P(J)+ P(S|J1)*P(J1)+ P(S|P)*P(P) =( 1 100 ∗ 0.6)+( 1 200 1 ∗ 0.2)+( 90 ∗ 0.15)+( 1 200 ∗ 0.05) =8.9167*10−3 P(J|S)= 𝑃(𝐽∩𝑆) 𝑝(𝑆) = 𝑃(𝑆|𝐽)∗𝑃(𝐽) = 1 ∗0.2 200 8.9167∗10−3 𝑃(𝑆) =0.1121 [12] From a box containing 5 black balls and 3 green balls, 2 balls are drawn in succession, the first ball being replaced in the box before the second draw is made. a- What is the probability that both balls are the same color?(with replacement) b- What is the probability that each color is represented? Solution: a- What is the probability that both balls are the same color? P(Both are the same color)=P(Both are black )+P(Both are green) 5 5 3 3 17 8 8 8 8 32 = ∗ + ∗ = b- What is the probability that each color is represented? P(Each color is represented)=P(1black,1green)+P(1green,1black) 5 3 3 5 15 8 8 8 8 32 = ∗ + ∗ = [13] Solve problem [12] in case of drawing the second ball without replacing the first one. Solution: a- What is the probability that both balls are the same color? P(Both are the same color)=P(Both are black )+P(Both are green) 5 4 3 2 13 8 7 8 7 28 = ∗ + ∗ = b- What is the probability that each color is represented? P(Each color is represented)=P(1black,1green)+P(1green,1black) 5 3 3 5 15 8 7 8 7 28 = ∗ + ∗ = [14] Multiple choices: 1. If the knowledge that an event A has occurred implies that a second event B cannot occur, the events A and B are said to be a- independent b- mutually exclusive c- A contains B d- non of the above. 2. If P(A)=0.4, P(B)=0.5 and A & B are disjoint, then the probability of occurrence of both A and B is: a- 0.0 b- 0.1 c- 0.2 d- 0.9 3. If P(A)=0.4, P(B)=0.45 and P (A𝐵̅ ) = 0.25 then P(AB) is : a- 0.7 b- 0.8 P(A ∩ B̅) = P(A) – P(A ∩ B)= 0.25 c- 0.9 d- non of the above. P(AUB) = P(A) + P(B) - P(A ∩ B) P(AUB) = 0.45+0.25 = 0.7 4. If P(A) = ½, P(B) = 1/3 and P(AB) = 2/3 then the events A and B are: a- independent b- mutually exclusive c- dependent d- non of the above. P(A ∩ B) = 1/2 + 1/3 - 2/3 = 1/6 , P(A ∩ B) = P(A) * P(B) = 1/6 5. If P (𝐴̅ ) = 0.6 and P (𝐴̅B ) = 0.5 then the probability of neither A nor B occurs is: a- 5/6 b- 0.1 c- 0.05 d- non of the above. P(A̅ ∩ B̅) = 1 - P(AUB) , P(A) = 1- 0.6 = 0.4 P(A̅ ∩ B) = P(B) - P(A ∩ B) = 0.5, p(AUB)=p(A)+P(B)- P(A ∩ B) P(AUB) = 0.4 + 0.5 = 0.9 P(A̅ ∩ B̅) = 1 – 0.9 = 0.1 6. If P(A)=P(B)=0.5 and P(AB)=0.75, then A and B are: a- Independent b- Mutually exclusive c- dependent d- A  B P(A ∩ B) = 0.5 + 0.5 – 0.75 = 0.25 P(A ∩ B) = P(A) * P(B) = 0.25 7. If P(A|B)=1, then A & B are: a- independent P(A/B) = P(A∩B) P(B) b- B  A = 1 , then P(A ∩ B) = P(B) c- A  B d- non of the above. 8. If P (A̅ | B ) = 1 , then A & B are: b- B  A a- independent (A̅ | B ) = c- A  B P(A̅∩B) P(B)− P(A∩B) P(B) P(B) = d- Mutually exclusive =1 , P(B) - P(A ∩ B) = P(B) 9. If P (B | A ) = 0 , then A & B are: a- independent b- B  A c- A  B d- Mutually exclusive 10. Event A occurs with probability 0.4., the conditional probability that A occurs given that B occurs is 0.5, while the conditional probability that A occurs given that B does not occur is 0.2. What is the conditional probability that B occurs given that A occurs? a- 5/6 given. b- 4/7 c- 5/8 d-Cannot be determined from the information