(A) Incorrect Recalculate the probabilities. Remember that ΣP(x) = 1. (B) Incorrect Recalculate the probabilities. Remember that ΣP(x) = 1.

Transcription

1 College Statistics - Problem Drill 11: Probability No. 1 of A sample space S consists of six simple events {A,B,C,D,E,F} with these probabilities P(A)=P(B)=0.10, P(C)=0.30, P(D)=2P(E), P(F)=0.20, and S 1 ={A,B,D}, S 2 ={E,F}. Which is correct below? (A) P(D)=0.10 (B) P(D)=0.30 (C) P(S 1 )=0.60 (D) P(S 2 )=0.30 (E) The simple event that are in both S 1 and S 2 is D. (A) Incorrect Recalculate the probabilities. Remember that ΣP(x) = 1. (B) Incorrect Recalculate the probabilities. Remember that ΣP(x) = 1. (C) Incorrect Recalculate the probabilities. Remember that ΣP(x) = 1. (D) Correct! Because P(A)+P(B)+P(C)+P(D)+P(E)+P(F)=1 and P(D)=2P(E), so p(E)+0.2=1 So P(E)=0.1 and then P(S 2 )=P(E)+P(F)= =0.3 (E) Incorrect! The answer is here. Recalculate the probabilities. Remember that ΣP(x) = 1. Applying the rules for probability, i.e. the sum total of the probability must be 1, because P(A)+P(B)+P(C)+P(D)+P(E)+P(F)=1 and P(D)=2P(E), p(E)+0.2=1 So P(E)=0.1 and P(S 2 )=P(E)+P(F)= =0.3 (D) P(S 2 )=0.30

2 No. 2 of 10 Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as 2. A glass jar contains 1 red, 1 green and 1 yellow marble, two marbles are chosen at random from the jar, one at a time and record their colors; decide which one is NOT true? (A) Sample space={rg,gr,ry,yr} (B) P(RG)=1/6 (C) P(YG)=1/6 (D) P(RY)=1/6 (E) If Sample space S 1 ={RG,RY}, then P(S 1 )=P(RG)+P(RY)=1/3 (A) Correct! This statement is not true. The sample space should be ={RG,GR,RY,YR,GY,YG} (B) Incorrect! This result is true, because sample space is = {RG,GR,RY,YR,GY,YG}, so it s obvious that P(RG)=1/6. (C) Incorrect! This result is true, because sample space is = {RG,GR,RY,YR,GY,YG}, so it s obvious that P(YG)=1/6. (D) Incorrect! This result is true because sample space is = {RG,GR,RY,YR,GY,YG}, so it s obvious that P(RY)=1/6. (E) Incorrect! This is true because sample space is = {RG,GR,RY,YR,GY,YG}, so it s obvious that P(S 1 )=P(RG)+P(RY)=1/3 Use Tree diagram to find all the sample space. The sample space should be ={RG,GR,RY,YR,GY,YG} Each event in the sample space has an equal chance of being chosen with a probability of 1/6. (A) Sample space={rg,gr,ry,yr}

3 No. 3 of 10 Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as 3. Use the counting rule to decide how many simple events are in the sample space. (A) When 5 coins are tossed, 2x5=10. (B) When 3 dices are tossed 3x6=18. (C) When 3 coins are tossed 2x2x2=8. (D) When 2 dices and 2 coins are tossed, 6x2=12. (E) When 3 dices are tossed, 6x3=18. A. Incorrect! When 5 coins are tossed, the following calculations define the number of simple events in the sample space: 2x2x2x2x2=32. B. Incorrect! When 3 die are tossed, 216 simple events are in the sample space (6x6x6=216). C. Correct! When 3 coins are tossed, 8 simple events are in the sample space; 2x2x2=8. D. Incorrect! When 2 die and 2 coins are tossed, 144 simple events are in the sample space: 6x6x2x2=144. E. Incorrect! When 3 die are tossed, 216 simple events are in the sample space.6x6x6=216. Use the counting rule, or (extended) mn rule to calculate. When 3 coins are tossed, 8 simple events are in the sample space; 2x2x2=8. (C) When 3 coins are tossed 2x2x2=8.

4 No. 4 of 10 Instructions: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as 4. Let A, B, C, D, E be simple events from an experiment, and event S, W and V are defined as follows: S={A,B}, P(S)=0.5, W={C,D} P(W)=0.3, V={B,C,D,E}, P(V)=0.8, find out the values for P(SUW) and P(S V) (A) P(SUW)=0.8 and P(S V)=0.3 (B) P(SUW)=0.3 and P(S V)=0.8 (C) P(SUW)=0.5 and P(S V)=0.3 (D) P(SUW)=0.8 and P(S V)=0.5 (E) P(SUW)=0.3 and P(S V)=0.5 (A) Correct! P(SUW)=P(S)+P(W)=0.8 P(A)=1-P(V)=1-0.8=0.2 P(S V)=P(B)=P(S)-P(A)= =0.3 (B) Incorrect! Calculate the following items to find the answer: P(SUW)=P(S)+P(W), P(A)=1-P(V), and P(S V)=P(B)=P(S)-P(A). (C) Incorrect! Calculate the following items to find the answer: P(SUW)=P(S)+P(W), P(A)=1-P(V), and P(S V)=P(B)=P(S)-P(A). (D) Incorrect! Calculate the following items to find the answer: P(SUW)=P(S)+P(W), P(A)=1-P(V), and P(S V)=P(B)=P(S)-P(A). (E) Incorrect! Calculate the following items to find the answer: P(SUW)=P(S)+P(W), P(A)=1-P(V), and P(S V)=P(B)=P(S)-P(A). Need to use the event relations which are union, intersection and complement. P(SUW)=P(S)+P(W)=0.8 P(A)=1-P(V)=1-0.8=0.2 P(S V)=P(B)=P(S)-P(A)= =0.3 (A) P(SUW)=0.8 and P(S V)=0.3

5 No. 5 of Similar to the previous question, let A,B,C,D,E be simple events from an experiment, and event S, W and V are defined as follows: S={A,B}, P(S)=0.5, W={C,D} P(W)=0.3, V={B,C,D,E}, P(V)=0.8. Find P(W V). (A) P(W V)=0.3 (B) P(W V)= 1.0 (C) P(W V)= 0.5 (D) P(W V)=.375 (E) P(W V)=.625 (A) Incorrect! P(W V)= P(W V)/P(V)=0.3/0.8=3/8. (B) Incorrect! P(W V)= P(W V)/P(V)=0.3/0.8=3/8. (C) Incorrect! P(W V)= P(W V)/P(V)=0.3/0.8=3/8. (D) Correct! P(W V)= P(W V)/P(V)=0.3/0.8=3/8. (E) Incorrect! P(W V)= P(W V)/P(V)=0.3/0.8=3/8. Need to use the event relations and the conditional probability: P(W V)=P(W)=0.3; P(W V)= P(W V)/P(V)=0.3/0.8=3/8; P(V W)= P(W V)/P(W)=0.3/0.3=1 (D) P(W V)=.375

6 No. 6 of If a sample space is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. A={3,6,9,12} are divisible by 3, so the probability is P(A)=4/12=1/3. Also B=(2,4,6,8,10,12) are divisible by 2, so the probability P(B)=1/2 What is the P(A B )? (A) P(A B )= 5/6 (B) P(A B )= 2/3 (C) P(A B )=1/6 (D) P(A B )= ½ (E) P(A B )= 1/4 A. Incorrect This is P(AUB), keep looking the correct answer is listed. B. Incorrect! The set of numbers divisible by both 2 and 3 is {6, 12}, The probability of this occurring is 2/12 or 1/6. C. Correct The set of numbers divisible by both 2 and 3 is {6, 12}, The probability of this occurring is 2/12 or 1/6. D. Incorrect The set of numbers divisible by both 2 and 3 is {6, 12}, The probability of this occurring is 2/12 or 1/6. E. Incorrect The set of numbers divisible by both 2 and 3 is {6, 12}, The probability of this occurring is 2/12 or 1/6. : The set of numbers divisible by both 2 and 3 is {6, 12}, The probability of this occurring is 2/12 or 1/6. (C) P(A B )=1/6

7 No. 7 of A teacher gave his class two tests, one in math and the other in statistics. Which is true? (A) 30% of the class passed both tests and 40% of the class passed the math test. What percent of those who passed the math test also passed the statistics test? P(statistics math)=3/4. (B) 70% of the class passed both tests and 80% of the class passed the math test. What percent of those who passed the math test also passed the statistics test? P(statistics math)=1/8. (C) 30% of the class passed both tests and 40% of the class passed the statistics test. What percent of those who passed the statistics test also passed the math test? P(math statistics)=1/4. (D) 70% of the class passed both tests and 80% of the class passed the statistics test. What percent of those who passed the statistics test also passed the math test? P(math statistics)=1/8. (E) 20% of the class passed both tests and 70% of the class passed the statistics test. What percent of those who passed the statistics test also passed the math test? P(math statistics)=5/7. A. Correct P(statistics math)= P(statistics math)/p(math) = 3/4 B. Incorrect! P(statistics math)= P(statistics math)/p(math) = 7/8 C. Incorrect! P(math statistics)= P(statistics math)/p(statistics) = 3/4 D. Incorrect! P(math statistics)= P(statistics math)/p(statistics) = 7/8 E. Incorrect P(math statistics)= P(statistics math)/p(statistics) = 2/7 A. P(statistics math)= P(statistics math)/p(math) = ¾ This is correct. B. P(statistics math)= P(statistics math)/p(math) = 7/8 C. P(math statistics)= P(statistics math)/p(statistics) = 3/4 D. P(math statistics)= P(statistics math)/p(statistics) = 7/8 E. P(math statistics)= P(statistics math)/p(statistics) = 2/7 (A) 30% of the class passed both tests and 40% of the class passed the math test. What percent of those who passed the math test also passed the statistics test? P(statistics math)=3/4.

8 No. 8 of True or false? IF two events C and D are independent,and P(C) =.2 and P(D) =.5, find P(C D). (A) P(C D)=.70 (B) P(C D)=.10 (C) P(C D)=.3 (D) P(C D)=.4 (E) P(C D)=.6 A. Incorrect. When two events are independent P(C D)=P(C)*P(D). B. Correct When two events are independent P(C D)=P(C)*P(D), = 2 *.5 =.10. C. Incorrect When two events are independent P(C D)=P(C)*P(D). D. Incorrect When two events are independent P(C D)=P(C)*P(D). E. Incorrect When two events are independent P(C D)=P(C)*P(D). The formula for probability of independent events is P(C D)=P(C)*P(D).. Since P(C)=.2 and P(D) =.5 P(C D)=.2 *.5 =.10 (B) P(C D)=.10

9 No. 9 of Coin-tossing and die-rolling are two independent events. When a coin is tossed once and a die is rolled twice. Find the probability of landing on the tail side of the coin and rolling a 4 and then a 3 on tossing the die twice. (A) (1/2)(1/6)(1/6) (B) 1/2+1/6+1/6 (C) (1/2)(1/6) (D) (1/6)(1/6) (E) 1/2+1/6+1/6-(1/2)(1/6)(1/6) A. Correct! (1/2)(1/6)(1/6) is correct. B. Incorrect! (1/2)(1/6)(1/6) is correct. C. Incorrect! (1/2)(1/6)(1/6) is correct. D. Incorrect! (1/2)(1/6)(1/6) is correct. E. Incorrect! (1/2)(1/6)(1/6) is correct. In independent events such as A, B,C,, we have P(A B C )=P(A)P(B)P(C), therefore (A) (1/2)(1/6)(1/6) is correct because these events are independent. (A) (1/2)(1/6)(1/6)

10 No. 10 of Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring. Which of the following examples is NOT true? (A) Landing on tails after tossing a coin and rolling a 2 on a die. (B) Choosing a marble from a jar and landing on tails after tossing a coin. (C) Rolling a 3 on a die, and then rolling a 3 on a second roll of the die. (D) A King is chosen at random from a standard deck of 52 cards. Then a second card is drawn, without replacement and it is a King. (E) Choosing a King from a deck of cards and rolling a 2 on a die. A. Incorrect! This is true because they are independent. B. Incorrect! This is true because they are independent. C. Incorrect! This is true because they are independent. D. Correct This is not true because they are not independent without well-reshuffling. E. Incorrect! This is true because they are independent. (A) Landing on tails after tossing a coin and rolling a 2 on a die. True because they are independent. (B) Choosing a marble from a jar and landing on tails after tossing a coin. True because they are independent. (C) Rolling a 3 on a die, and then rolling a 3 on a second roll of the die. True because they are independent. (D) A King is chosen at random from a standard deck of 52 cards. Then without replacement a second card is drawn and it s a King again. These are not independent events because the probability of getting a king changes from the first draw to the 2 nd. (E) These two events are independent. This statement is true.