DAT - Dental Admission Quantitative Reasoning: Probability Questions and Answers 1 — Questions and Answers
Question 1: A dental clinic uses two independent sterilization machines, A and B. Machine A has a 95% success rate, and Machine B has a 90% success rate. If a set of instruments is processed through both machines, what is the probability that at least one machine successfully sterilizes the set?
- 0.855
- 0.950
- 0.995 (Correct answer)
- 0.975
Correct answer: 0.995
To find the probability of at least one success, it's easiest to first calculate the probability of the complementary event, which is that both machines fail. Then, subtract this from 1. Probability of Machine A failing = 1 - 0.95 = 0.05. Probability of Machine B failing = 1 - 0.90 = 0.10. Since the events are independent, the probability of both failing is the product of their individual probabilities: 0.05 * 0.10 = 0.005. The probability of at least one machine succeeding is 1 - P(both fail) = 1 - 0.005 = 0.995.
Question 2: A dental study group consists of 5 first-year students and 4 second-year students. If a committee of 4 students is selected at random, what is the probability that the committee will consist of exactly 2 first-year students and 2 second-year students?
- 10/21 (Correct answer)
- 1/2
- 5/14
- 25/126
Correct answer: 10/21
First, calculate the total number of ways to choose a committee of 4 from the 9 students using combinations: C(9,4) = 9! / (4! * 5!) = (9*8*7*6)/(4*3*2*1) = 126 total possible committees. Next, calculate the number of ways to choose 2 first-year students from 5: C(5,2) = 5! / (2! * 3!) = 10. Then, calculate the number of ways to choose 2 second-year students from 4: C(4,2) = 4! / (2! * 2!) = 6. The number of ways to form the specific committee is the product of these two results: 10 * 6 = 60 favorable outcomes. The probability is the ratio of favorable outcomes to total outcomes: 60/126, which simplifies to 10/21.
Question 3: In a standard 52-card deck, what is the probability of drawing a single card that is a King or a Spade?
- 17/52
- 4/13 (Correct answer)
- 1/4
- 1/13
Correct answer: 4/13
This problem involves non-mutually exclusive events because a card can be both a King and a Spade (the King of Spades). The formula is P(A or B) = P(A) + P(B) - P(A and B). P(King) = 4/52. P(Spade) = 13/52. P(King and Spade) = 1/52. Therefore, P(King or Spade) = (4/52) + (13/52) - (1/52) = 16/52. This fraction simplifies to 4/13.
Question 4: A bag contains 6 red marbles, 4 blue marbles, and 5 yellow marbles. If one marble is drawn at random, which of the following events has the highest probability of occurring?
- Drawing a red marble
- Drawing a blue marble
- Drawing a marble that is not red
- Drawing a marble that is not yellow (Correct answer)
Correct answer: Drawing a marble that is not yellow
First, find the total number of marbles: 6 + 4 + 5 = 15. Now, calculate the probability of each event: A) P(Red) = 6/15 = 0.40 B) P(Blue) = 4/15 ≈ 0.27 C) P(Not Red) = P(Blue or Yellow) = (4+5)/15 = 9/15 = 0.60 D) P(Not Yellow) = P(Red or Blue) = (6+4)/15 = 10/15 ≈ 0.67 Comparing the probabilities, drawing a marble that is not yellow (0.67) has the highest probability.
Question 5: A fair six-sided die is rolled three times. What is the probability of rolling a '4' on the first roll, an even number on the second roll, and a number greater than 2 on the third roll?
- 1/12
- 1/3
- 1/18 (Correct answer)
- 1/36
Correct answer: 1/18
These are three independent events, so their combined probability is the product of their individual probabilities. 1. Probability of rolling a '4' = 1/6. 2. Probability of rolling an even number ({2, 4, 6}) = 3/6 = 1/2. 3. Probability of rolling a number greater than 2 ({3, 4, 5, 6}) = 4/6 = 2/3. Multiply the probabilities together: (1/6) * (1/2) * (2/3) = 2/36 = 1/18.
Question 6: A survey of 100 dental patients reveals that 60 floss daily and 35 have at least one cavity. Of the patients who floss daily, 15 have a cavity. What is the probability that a randomly selected patient has at least one cavity, given that they do not floss daily?
- 3/8
- 7/20
- 1/4
- 1/2 (Correct answer)
Correct answer: 1/2
This is a conditional probability problem. Let C be the event of having a cavity and F' be the event of not flossing daily. We want to find P(C|F'). First, find the number of patients in each category: Total patients = 100. Floss daily = 60, so Do not floss daily = 100 - 60 = 40. Total with a cavity = 35. Floss daily AND have a cavity = 15. This means the number of patients who do not floss daily AND have a cavity is: (Total with cavity) - (Flossers with cavity) = 35 - 15 = 20. The conditional probability P(C|F') is the number of patients who don't floss and have a cavity, divided by the total number of patients who don't floss. P(Cavity | Do not floss) = 20 / 40 = 1/2.
DAT - Dental Admission Quantitative Reasoning: Probability Questions and Answers 1
A dental clinic uses two independent sterilization machines, A and B.
Machine A has a 95% success rate, and Machine B has a 90% success rate.
If a set of instruments is processed through both machines, what is the probability that at least one machine successfully sterilizes the set?