Probability review Problem set B One question at a time

Two fair coins are tossed one after the other.

1. (a) Given that at least one is a head, find the probability that both are heads.

   • Sample space (first coin first): HH, HT, TH, TT
   • At least one head rules out TT. Left: HH, HT, TH
   • Only HH has two heads: \(P=\frac{1}{3}\)

2. (b) Given that the first coin is a head, find the probability that both are heads.

   • First is a head: HH and HT are left
   • \(P=\frac{1}{2}\)

A school bus is cancelled on any given day with probability 0.06, independently of other days.

1. (a) Find the probability that the bus is cancelled on at least one of 3 days.

   • \(P(\text{none})=0.94^{3}=0.831\)
   • \(P(\text{at least one})=1-0.831=0.169\)

2. (b) Find the probability that the bus is cancelled on all of 3 days.

   • Multiply: \(0.06^{3}=0.000216\)

A raffle bag holds 10 red tickets and 4 blue tickets. Two tickets are drawn at random one after the other, without replacement.

1. (a) Given that the first is red, find the probability that the second is also red.

   • After one red is taken: 9 red left out of 13.
   • \(P=\frac{9}{13}\)

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

   • \(P=\frac{10}{14}\times\frac{9}{13}=\frac{45}{91}\)

3. (c) Is it more likely that the two are the same colour or different colours? Justify your answer.

   • Both red: \(\frac{45}{91}\). Both blue: \(\frac{4}{14}\times\frac{3}{13}=\frac{6}{91}\).
   • Same: \(\frac{51}{91}\). Different: \(1-\frac{51}{91}=\frac{40}{91}\)
   • \(\frac{51}{91}\) is larger, so same is more likely.

Of 90 visitors to a town, 36 went to the museum, 40 to the zoo and 30 to the park. 12 went to the museum and the zoo, 10 to the zoo and the park, 9 to the museum and the park, and 4 went to all three.

1. (a) Draw a Venn diagram and fill in every region, including the region outside the circles.

   • Start in the middle: 4 in all three.
   • Two only: M and Z only: 12 − 4 = 8. Z and P only: 10 − 4 = 6. M and P only: 9 − 4 = 5.
   • One only: M only 19, Z only 22, P only 15.
   • Outside: 90 − 79 = 11

2. (b) Find the probability that a person chosen at random went to the park only.

   • \(P=\frac{15}{90}=\frac{1}{6}\)

3. (c) Given that a person chosen at random went to the park, find the probability that they also went to the zoo.

   • Circle P holds 30 people. 10 of them are in Z too.
   • \(P=\frac{10}{30}=\frac{1}{3}\)

Each of 35 students takes at least one of art (A), music (M) or drama (D). 16 take art, 19 take music and 17 take drama. 5 take art and music, 8 take music and drama, and 7 take art and drama.

1. (a) Let x be the number of students who take all three subjects. Draw a Venn diagram showing this information, with each region in terms of x.

   • All three: x. Two only: take x from each overlap: 5 − x, 8 − x, 7 − x.
   • One only: circle total, minus the two overlaps, plus x (taken off twice). A only: 16 − 5 − 7 + x = 4 + x.
   • M only: 6 + x. D only: 2 + x.

2. (b) Show that \(x+32=35\).

   • Add all seven regions. The x terms give \(1-3+3=1\) lots of x.
   • The numbers add to 32, so the total is \(x+32\).
   • Everyone has at least one item, so \(x+32=35\).

3. (c) Find the probability that a student chosen at random has all three.

   • \(x=35-32=3\)
   • \(P=\frac{3}{35}\)

4. (d) Given that a student does not take drama, find the probability that they take art.

   • Outside circle D: A only 7, M only 9, A and M only 2, total 18.
   • Of these, 9 are in circle A: \(P=\frac{9}{18}=\frac{1}{2}\)

5. (e) Given that a student has exactly two of the three, find the probability that they take music and drama.

   • Exactly two: 2 + 5 + 4 = 11.
   • \(P=\frac{5}{11}\)

Three fair coins are tossed one after another. Write H for a head and T for a tail.

1. (a) List the 8 equally likely outcomes.

   • Write each outcome in order: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

2. (b) Given that at least one is a head, find the probability that all three are heads.

   • At least one is a head rules out only the outcome with none of that kind. 7 outcomes are left.
   • Exactly one of them has all three the same kind.
   • \(P=\frac{1}{7}\)

3. (c) Given that the first coin is a head, find the probability that all three are heads.

   • The first coin is a head: the 4 outcomes starting with that kind are left.
   • One of those 4 has all three the same kind.
   • \(P=\frac{1}{4}\)

Two machines in a factory break down on a given day with probabilities 0.15 and 0.25, independently. Find the probability that at least one of them breaks down on a given day.

• Independent, so \(P(\text{neither})=(1-0.15)(1-0.25)=0.6375\)
• \(P(\text{at least one})=1-0.6375=0.3625\)