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IB Mathematics: Applications and Interpretation SL · 13.1: Probability · 13.1.6

Conditional probability Problem set B Worked solutions

Skills

  1. A box holds 7 chocolates and 5 mints. Two sweets are taken out at random, one after the other, without replacement. The first sweet is a mint.

    1. (a) Find the probability that the second sweet is a mint.

      • The first sweet has gone, so 11 sweets are left
      • Of these, 4 are mints
      • \(P=\frac{4}{11}\)

    2. (b) Find the probability that the second sweet is a chocolate.

      • 11 sweets are left
      • Of these, 7 are chocolates
      • \(P=\frac{7}{11}\)

  2. A game uses 36 cards: four suits of nine cards, numbered 1 to 9. Two cards are taken at random, without replacement. The first card is the 7 of hearts.

    1. (a) Find the probability that the second card is a heart.

      • The 7 of hearts has gone, so 35 cards are left
      • 8 of them are in the same suit
      • \(P=\frac{8}{35}\)

    2. (b) Find the probability that the second card is a 7.

      • 35 cards are left
      • 3 of them have the same value as the first card
      • \(P=\frac{3}{35}\)

  3. The table shows the type of film and the age group of the people in a cinema one evening.

    FilmUnder 1818 or overComedy2431Thriller933Cartoon2815

    1. (a) Find the probability that a person chosen at random is under 18, given that the person watches a thriller.

      • \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\)
      • \(P(A\cap B)=\frac{9}{140}\), \(P(B)=\frac{42}{140}\)
      • \(P=\frac{9}{42}=\frac{3}{14}\)

    2. (b) Find the probability that a person chosen at random watches a cartoon, given that the person is 18 or over.

      • \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\)
      • \(P(A\cap B)=\frac{15}{140}\), \(P(B)=\frac{79}{140}\)
      • \(P=\frac{15}{79}\)

  4. In a group of 80 members of a club, 44 cycle to the club, 30 come by bus and 14 do both on different days. Let C be the set who cycle and B the set who come by bus. Draw a Venn diagram.

    1. (a) Find the probability that a member comes by bus, given that the member cycles.

      • Draw the Venn diagram: both 14; only C 44 − 14 = 30; only B 30 − 14 = 16; neither 80 − (30 + 14 + 16) = 20
      • Given: cycles. Keep only these regions: 30 + 14 = 44
      • Of these, comes by bus: 14
      • \(P=\frac{14}{44}=\frac{7}{22}\)

    2. (b) Find the probability that a member cycles, given that the member does not come by bus.

      • Given: does not come by bus. Keep only these regions: 30 + 20 = 50
      • Of these, cycles: 30
      • \(P=\frac{30}{50}=\frac{3}{5}=0.6\)

  5. The Venn diagram shows how 50 children use R, the roller rink, and G, the games room.

    1781213RG

    1. (a) Find the probability that a child uses the games room, given that the child uses the rink.

      • Given: uses the rink. Keep only these regions: 17 + 8 = 25
      • Of these, uses the games room: 8
      • \(P=\frac{8}{25}=0.32\)

    2. (b) Find the probability that a child uses neither, given that the child does not use the rink.

      • Given: does not use the rink. Keep only these regions: 12 + 13 = 25
      • Of these, does neither: 13
      • \(P=\frac{13}{25}=0.52\)

In context

  1. A survey of 90 teenagers found that 50 own a bike, 35 own a scooter and 15 own both. Let B be the set who own a bike and S the set who own a scooter.

    1. (a) Find the probability that a teenager owns a scooter, given that the teenager owns a bike.

      • Draw the Venn diagram: both 15; only B 50 − 15 = 35; only S 35 − 15 = 20; neither 90 − (35 + 15 + 20) = 20
      • Given: owns a bike. Keep only these regions: 35 + 15 = 50
      • Of these, owns a scooter: 15
      • \(P=\frac{15}{50}=\frac{3}{10}=0.3\)

    2. (b) Find the probability that a teenager owns a bike, given that the teenager owns no scooter.

      • Given: owns no scooter. Keep only these regions: 35 + 20 = 55
      • Of these, owns a bike: 35
      • \(P=\frac{35}{55}=\frac{7}{11}\)

  2. A customer at a café is chosen at random. The probability that the customer buys a drink but no food is 0.24. The probability that the customer buys both is 0.18. The probability that the customer buys neither is 0.36.

    1. (a) Find the probability that the customer buys food, given that the customer buys a drink.

      • Given: buys a drink. Keep only these regions: 0.24 + 0.18 = 0.42
      • Of these, buys food: 0.18
      • \(P=\frac{0.18}{0.42}=\frac{3}{7}\)

    2. (b) Find the probability that the customer buys a drink, given that the customer buys food.

      • Given: buys food. Keep only these regions: 0.18 + 0.22 = 0.4
      • Of these, buys a drink: 0.18
      • \(P=\frac{0.18}{0.4}=\frac{9}{20}=0.45\)

Stretch

  1. A drawer holds 6 grey, 3 white and 2 black socks. Two socks are taken out at random, one after the other, without replacement.

    1. (a) Given that the first sock is white, find the probability that the second sock is not white.

      • A white sock has gone: 10 socks left
      • Not white: 11 − 3 = 8 (none of these has been taken)
      • \(P=\frac{8}{10}=\frac{4}{5}=0.8\)

    2. (b) Find the probability that both socks are grey.

      • First grey: \(\frac{6}{11}\)
      • Then, given that, second grey: \(\frac{5}{10}\)
      • \(P(\text{both})=\frac{6}{11}\times\frac{5}{10}=\frac{3}{11}\)

  2. At a leisure centre, A is the event that a visitor swims and B is the event that the visitor uses the gym. P(A) = 0.7, P(B) = 0.5 and P(A ∪ B) = 0.85.

    1. (a) Find P(A ∩ B).

      • \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
      • \(0.85=0.7+0.5-P(A\cap B)\)
      • \(P(A\cap B)=0.7+0.5-0.85=0.35\)

    2. (b) Find P(A | B).

      • \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\)
      • \(=\frac{0.35}{0.5}=\frac{7}{10}=0.7\)

jmaths.xyz · AI SL · 13.1.6 Conditional probability