Conditional probability Problem set A One question at a time

A bag contains 9 yellow counters and 6 green counters. Two counters are taken out at random, one after the other, without replacement. The first counter is yellow.

1. (a) Find the probability that the second counter is yellow.

   • The first counter has gone, so 14 counters are left
   • Of these, 8 are yellow
   • \(P=\frac{8}{14}=\frac{4}{7}\)

2. (b) Find the probability that the second counter is green.

   • 14 counters are left
   • Of these, 6 are green
   • \(P=\frac{6}{14}=\frac{3}{7}\)

Two cards are taken at random from a standard pack of 52 cards, without replacement. The first card is the ace of clubs.

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

   • The ace of clubs has gone, so 51 cards are left
   • 12 of them are in the same suit
   • \(P=\frac{12}{51}=\frac{4}{17}\)

2. (b) Find the probability that the second card is an ace.

   • 51 cards are left
   • 3 of them have the same value as the first card
   • \(P=\frac{3}{51}=\frac{1}{17}\)

The table shows the subject chosen and the gender of the students in a sixth form.

1. (a) Find the probability that a student chosen at random is a girl, given that the student takes biology.

   • \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\)
   • \(P(A\cap B)=\frac{27}{100}\), \(P(B)=\frac{45}{100}\)
   • \(P=\frac{27}{45}=\frac{3}{5}=0.6\)

2. (b) Find the probability that a student chosen at random takes art, given that the student is a boy.

   • \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\)
   • \(P(A\cap B)=\frac{6}{100}\), \(P(B)=\frac{45}{100}\)
   • \(P=\frac{6}{45}=\frac{2}{15}\)

In a group of 60 students, 35 play netball, 28 play hockey and 12 play both. Let N be the set who play netball and H the set who play hockey. Draw a Venn diagram.

1. (a) Find the probability that a student plays hockey, given that the student plays netball.

   • Draw the Venn diagram: both 12; only N 35 − 12 = 23; only H 28 − 12 = 16; neither 60 − (23 + 12 + 16) = 9
   • Given: plays netball. Keep only these regions: 23 + 12 = 35
   • Of these, plays hockey: 12
   • \(P=\frac{12}{35}\)

2. (b) Find the probability that a student plays netball, given that the student does not play hockey.

   • Given: does not play hockey. Keep only these regions: 23 + 9 = 32
   • Of these, plays netball: 23
   • \(P=\frac{23}{32}=0.71875\)

The Venn diagram shows how 40 members of a club use T, the tennis courts, and S, the squash courts.

1. (a) Find the probability that a member uses squash, given that the member uses tennis.

   • Given: uses tennis. Keep only these regions: 14 + 11 = 25
   • Of these, uses squash: 11
   • \(P=\frac{11}{25}=0.44\)

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

   • Given: does not use tennis. Keep only these regions: 9 + 6 = 15
   • Of these, does neither: 6
   • \(P=\frac{6}{15}=\frac{2}{5}=0.4\)

A survey of 120 households found that 66 have a garden, 48 have a pet and 30 have both. Let G be the set with a garden and P the set with a pet.

1. (a) Find the probability that a household has a pet, given that it has a garden.

   • Draw the Venn diagram: both 30; only G 66 − 30 = 36; only P 48 − 30 = 18; neither 120 − (36 + 30 + 18) = 36
   • Given: has a garden. Keep only these regions: 36 + 30 = 66
   • Of these, has a pet: 30
   • \(P=\frac{30}{66}=\frac{5}{11}\)

2. (b) Find the probability that a household has a garden, given that it has no pet.

   • Given: has no pet. Keep only these regions: 36 + 36 = 72
   • Of these, has a garden: 36
   • \(P=\frac{36}{72}=\frac{1}{2}=0.5\)

A student at a college is chosen at random. The probability that the student runs but does not swim is 0.30. The probability that the student runs and swims is 0.15. The probability that the student does neither is 0.30.

1. (a) Find the probability that the student swims, given that the student runs.

   • Given: runs. Keep only these regions: 0.3 + 0.15 = 0.45
   • Of these, swims: 0.15
   • \(P=\frac{0.15}{0.45}=\frac{1}{3}\)

2. (b) Find the probability that the student runs, given that the student swims.

   • Given: swims. Keep only these regions: 0.15 + 0.25 = 0.4
   • Of these, runs: 0.15
   • \(P=\frac{0.15}{0.4}=\frac{3}{8}=0.375\)

A bag holds 5 red, 4 blue and 3 green counters. Two counters are taken out at random, one after the other, without replacement.

1. (a) Given that the first counter is blue, find the probability that the second counter is not blue.

   • A blue counter has gone: 11 counters left
   • Not blue: 12 − 4 = 8 (none of these has been taken)
   • \(P=\frac{8}{11}\)

2. (b) Find the probability that both counters are red.

   • First red: \(\frac{5}{12}\)
   • Then, given that, second red: \(\frac{4}{11}\)
   • \(P(\text{both})=\frac{5}{12}\times\frac{4}{11}=\frac{5}{33}\)

In a village, A is the event that a household has broadband and B is the event that it has a landline. P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.65.

1. (a) Find P(A ∩ B).

   • \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
   • \(0.65=0.5+0.4-P(A\cap B)\)
   • \(P(A\cap B)=0.5+0.4-0.65=0.25\)

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

   • \(P(A\mid B)=\frac{P(A\cap B)}{P(B)}\)
   • \(=\frac{0.25}{0.4}=\frac{5}{8}=0.625\)