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Metal Detector Probability Calculations

  • Scenario: James uses a homemade metal detector to locate valuable objects at the beach.

  • Detector Efficacy:

    • Accuracy: Beeps for 98% of metal objects.
    • False Positives: Beeps for 4% of non-metal objects.
    • Metal Object Proportion: 25% of the objects detected are metal.

Probability Questions:

  • 1. Probability James passes over a metal item AND the detector beeps:

    • Calculation:
    • Let M be the event that James passes over a metal item, P(M) = 0.25.
    • Let B be the event that the detector beeps given that a metal item is present; P(B|M) = 0.98.
    • Using the multiplication rule:
      P(MextANDB)=P(M)imesP(B∣M)=0.25imes0.98=0.245.P(M ext{ AND } B) = P(M) imes P(B|M) = 0.25 imes 0.98 = 0.245.
    • Answer: 0.245.

  • 2. Probability James passes over a non-metal item AND the detector beeps (false alarm):

    • Calculation:
    • Let N be the event that James passes over a non-metal item, P(N) = 0.75 (since 100% - 25% = 75%).
    • Let B be the event that the detector beeps given a non-metal item; P(B|N) = 0.04.
    • Using the multiplication rule:
      P(NextANDB)=P(N)imesP(B∣N)=0.75imes0.04=0.03.P(N ext{ AND } B) = P(N) imes P(B|N) = 0.75 imes 0.04 = 0.03.
    • Answer: 0.03.

  • 3. Probability that the detector beeps:

    • Total Probability of Detector Beeping:
    • Use the Law of Total Probability:
      P(B)=P(M)imesP(B∣M)+P(N)imesP(B∣N)P(B) = P(M) imes P(B|M) + P(N) imes P(B|N)
      =0.25imes0.98+0.75imes0.04= 0.25 imes 0.98 + 0.75 imes 0.04
      =0.245+0.03=0.275.= 0.245 + 0.03 = 0.275.
    • Answer: 0.275.

  • 4. Given that the detector beeps, the probability that it passed over a metal item:

    • Calculation using Bayes' theorem:
      P(M∣B)=P(MextANDB)P(B)=0.2450.275≈0.891.P(M|B) = \frac{P(M ext{ AND } B)}{P(B)} = \frac{0.245}{0.275} \approx 0.891.
    • Answer: 0.891.

Bag of Marbles Probability Calculations

  • Scenario: A bag of marbles contains:
    • 5 Red Marbles
    • 7 Blue Marbles
    • 8 Black Marbles
    • Total Number of Marbles: 5 + 7 + 8 = 20.

Probability Questions:

  • 1. Probability of selecting a red marble and then selecting a black marble (without replacement):

    • Calculation:
    • Probability of selecting a red marble first:
      P(R)=520=14.P(R) = \frac{5}{20} = \frac{1}{4}.
    • After selecting a red marble, there are now 19 marbles left (4 red, 7 blue, 8 black):
    • Probability of selecting a black marble afterwards:
      P(B∣R)=819.P(B|R) = \frac{8}{19}.
    • Therefore, the answer is:
      P(RextandB)=P(R)imesP(B∣R)=520×819.P(R ext{ and } B) = P(R) imes P(B|R) = \frac{5}{20} \times \frac{8}{19}.

  • 2. Probability of selecting 3 blue marbles in a row (without replacement):

    • Calculation:
    • Probability of selecting the first blue marble:
      P(B1)=720.P(B_1) = \frac{7}{20}.
    • Second blue marble, now 6 blue marbles left:
      P(B<em>2∣B</em>1)=619.P(B<em>2|B</em>1) = \frac{6}{19}.
    • Third blue marble, now 5 blue marbles left:
      P(B<em>3∣B</em>1extandB2)=518.P(B<em>3|B</em>1 ext{ and } B_2) = \frac{5}{18}.
    • Therefore, the final expression is:
      P(B<em>1extandB</em>2extandB<em>3)=P(B</em>1)imesP(B<em>2∣B</em>1)imesP(B<em>3∣B</em>1extandB2)=720×619×518.P(B<em>1 ext{ and } B</em>2 ext{ and } B<em>3) = P(B</em>1) imes P(B<em>2|B</em>1) imes P(B<em>3|B</em>1 ext{ and } B_2) = \frac{7}{20} \times \frac{6}{19} \times \frac{5}{18}.

  • 3. Probability of selecting 1 marble at a time (without replacement) and not getting a red one until the 4th pick:

    • Calculation:
    • First three picks need to be non-red (blue or black), which from the beginning:
    • Probability of first pick not being red:
      P(NR1)=1520.P(NR_1) = \frac{15}{20}.
    • Second pick not red (14 left):
      P(NR<em>2∣NR</em>1)=1419.P(NR<em>2|NR</em>1) = \frac{14}{19}.
    • Third pick not red (13 left):
      P(NR<em>3∣NR</em>1extandNR2)=1318.P(NR<em>3|NR</em>1 ext{ and } NR_2) = \frac{13}{18}.
    • Fourth pick must be red:
      P(R∣NR<em>1extandNR</em>2extandNR3)=517.P(R|NR<em>1 ext{ and } NR</em>2 ext{ and } NR_3) = \frac{5}{17}.
    • Therefore, the final expression is:
      P(NR<em>1extandNR</em>2extandNR3extandR)=1520×1419×1318×517.P(NR<em>1 ext{ and } NR</em>2 ext{ and } NR_3 ext{ and } R) = \frac{15}{20} \times \frac{14}{19} \times \frac{13}{18} \times \frac{5}{17}.

Basic Multiplication Rule

  • Definition: If A and B are independent events, the probability of both A and B occurring:
    P(AextandB)=P(A)imesP(B).P(A ext{ and } B) = P(A) imes P(B).

Probability Questions:

  • 1. Probability of tossing a coin and getting heads, and rolling a die to get a 5:

    • Use the multiplication rule:
      P(Hextand5)=P(H)imesP(5)=12×16=112.P(H ext{ and } 5) = P(H) imes P(5) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}.

  • 2. Probability that you toss a coin 5 times and get all 5 heads:

    • Probability:
      P(HHHHH)=(12)5=132.P(HHHHH) = \left(\frac{1}{2}\right)^5 = \frac{1}{32}.

  • 3. If P(Type O blood) = 0.44, the probability that the first 3 donors at a blood drive are Type O:

    • Calculation:
      P(O<em>1extandO</em>2extandO3)=P(O)3=0.443=0.085184.P(O<em>1 ext{ and } O</em>2 ext{ and } O_3) = P(O)^3 = 0.44^3 = 0.085184.

  • 4. If P(O) = 0.44, what is the probability that the first Type O donor is the 4th donor:

    • This follows a geometric distribution:
      P(N=4)=P(NR<em>1)imesP(NR</em>2)imesP(NR3)imesP(O)=(1−0.44)3imes0.44=0.563imes0.44.P(N = 4) = P(NR<em>1) imes P(NR</em>2) imes P(NR_3) imes P(O) = (1 - 0.44)^3 imes 0.44 = 0.56^3 imes 0.44.

Complement Rule

  • Definition: Complement rule states that the probability of an event happening at least once is:
    • At least once: 1 - **never.

Probability Questions:

  • 1. If you roll a die 4 times, the probability you get at least one 1:

    • Calculation:
      P(atleastoneext1)=1−P(noext1s)=1−(56)4.P(at least one ext{ } 1) = 1 - P(no ext{ } 1s) = 1 - \left(\frac{5}{6}\right)^4.

  • 2. If P(O) = 0.44, the probability there is at least one Type O person among the first 5 donors:

    • Calculation:
      P(atleastoneextO)=1−P(noextO)=1−(1−0.44)5=1−0.565.P(at least one ext{ } O) = 1 - P(no ext{ } O) = 1 - (1 - 0.44)^5 = 1 - 0.56^5.

  • 3. If the probability of passing a certain test is 0.15, what is the probability at least one student in a class of 15 passes:

    • Calculation:
      P(atleastoneextpass)=1−P(noextpasses)=1−(0.85)15.P(at least one ext{ } pass) = 1 - P(no ext{ } passes) = 1 - (0.85)^{15}.

General Multiplication Rule

  • Definition: The general multiplication rule for two events A and B is:
    P(AextandB)=P(A)imesP(B∣A).P(A ext{ and } B) = P(A) imes P(B | A).

Counting Principle

  • Rule:
    • OR adds outcomes.
    • AND multiplies outcomes.

Counting Questions:

  • 1. Possible outcomes for 5 coin tosses:

    • Calculation:
      25=322^5 = 32 outcomes.

  • 2. Possible phone numbers in the 561 area code that start with a 6 or 7:

    • Calculation: Two conditions:
    • Starts with 6: 561-6xx-xxxx: 10 options for each x 10 x 10 = 1000
    • Starts with 7: 561-7xx-xxxx: same calculation as above.
    • Total possible combinations: 1000 + 1000 = 2000.

  • 3. Early bird dinner special combinations:

    • Main dish options = 5,
    • Dessert options = 3,
    • Drink options = 4.
    • Total combinations:
      5imes3imes4=60.5 imes 3 imes 4 = 60.

Permutations

  • Definition: If the arrangement of outcomes matters, use the permutations formula.

Permutation Questions:

  • 1. Ways to choose a specific playlist of 4 songs from a 10-song playlist:

    • Calculation:
      P(10,4)=10!(10−4)!=10!6!=10×9×8×7.P(10,4) = \frac{10!}{(10-4)!} = \frac{10!}{6!} = 10 \times 9 \times 8 \times 7.

  • 2. With 12 runners in a race, the arrangements of 1st, 2nd, and 3rd place:

    • Calculation:
      P(12,3)=12!(12−3)!=12!9!=12×11×10.P(12,3) = \frac{12!}{(12-3)!} = \frac{12!}{9!} = 12 \times 11 \times 10.

  • 3. 3 letter words from the word FACTOR:

    • Each letter can be chosen for each position without repetition:
      P(6,3)=6×5×4.P(6,3) = 6 \times 5 \times 4.

  • 4. A club with 16 members choosing a President, Vice-President, and Secretary:

    • Calculation:
      P(16,3)=16×15×14.P(16,3) = 16 \times 15 \times 14.

Combinations

  • Definition: If order does not matter, use combinations formula.

Combination Questions:

  • 1. How many different 5-card hands from a deck of 52 cards:

    • Calculation:
      C(52,5)=52!5!(52−5)!=52!5!47!.C(52,5) = \frac{52!}{5!(52-5)!} = \frac{52!}{5!47!}.

  • 2. Craft items from 9 different items, choosing 4:

    • Calculation:
      C(9,4)=9!4!(9−4)!=9!4!5!.C(9,4) = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!}.

  • 3. 12 runners in a race choosing 3 to advance:

    • Calculation:
      C(12,3)=12!3!(12−3)!=12!3!9!.C(12,3) = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!}.

  • 4. Club with 20 members forming a 5-person committee:

    • Calculation:
      C(20,5)=20!5!(20−5)!=20!5!15!.C(20,5) = \frac{20!}{5!(20-5)!} = \frac{20!}{5!15!}.

Mixed Practice & Probability Questions

  • 1. Lottery probability needing to match 5 different numbers from 1-40:

    • Total combinations = 40 choose 5:
      C(40,5)=40!5!(40−5)!.C(40,5) = \frac{40!}{5!(40-5)!}.

  • 2. 2 oldest people chosen among 6 board members for a special project:

    • Probability:
      P(2extoldest)=C(2,2)×C(4,0)C(6,2).P(2 ext{ oldest}) = \frac{C(2,2) \times C(4,0)}{C(6,2)}.

  • 3. Winning a lottery needing to choose 4 numbers in correct order:

    • Probability:
      P(4extcorrect)=120!/(20−4)!=120×19×18×17.P(4 ext{ correct}) = \frac{1}{20!/(20-4)!} = \frac{1}{20 \times 19 \times 18 \times 17}.

  • 4. Tossing a coin 3 times and getting exactly 1 tail:

    • Calculation:
      P(1exttailin3tosses)=C(3,1)×(12)3=3×18=38.P(1 ext{ tail in 3 tosses}) = C(3,1) \times \left(\frac{1}{2}\right)^3 = 3 \times \frac{1}{8} = \frac{3}{8}.

  • 5. Probability all 4 on committee are Republicans from 8 Democrats, 6 Republicans, and 2 Independents:

    • Calculation:
      P(all4Republicans)=C(6,4)C(16,4).P(all 4 Republicans) = \frac{C(6,4)}{C(16,4)}.

Quiz Questions

  • 1. Pizza combinations from options:

    • Crust options = 2,
    • Sauce options = 4,
    • Topping options = 5.
    • Total combinations:
      2×4×5=40.2 \times 4 \times 5 = 40.

  • 2. Possibilities for a 6-digit ATM PIN code using digits 0-9:

    • Calculation:
      106=1,000,000.10^6 = 1,000,000.