Tree diagrams and sample spaces Problem set A One question at a time

Every day, a student either walks to school (probability 0.6) or goes by bus. If the student walks, the probability of being late is 0.1. If the student goes by bus, the probability of being late is 0.25. Find the probability that the student

1. (a) walks and is late.

   • Draw the tree: first stage walk or not walk.
   • 0.6 × 0.1 = 0.06

2. (b) is late.

   • The other path: 0.4 × 0.25 = 0.1
   • Add the two paths: 0.06 + 0.1 = 0.16

A bag contains 4 yellow counters and 6 green counters. Two counters are taken at random, one after the other, without replacement. Find the probability that

1. (a) both are yellow.

   • \(P(\text{yellow}\text{ then }\text{yellow})=\frac{4}{10}\times\frac{3}{9}=\frac{2}{15}\)

2. (b) the two are different colours.

   • \(P(\text{yellow then green})=\frac{4}{10}\times\frac{6}{9}=\frac{4}{15}\)
   • The other order gives the same probability.
   • \(P=2\times\frac{4}{15}=\frac{8}{15}\)

The Venn diagram shows probabilities for the students at a school. P is the event "has a phone" and T is the event "has a tablet". A student is chosen at random.

1. (a) Find the value of x.

   • The probabilities add to 1: 0.32 + 0.18 + 0.27 + x = 1
   • x = 0.23

2. (b) Find the probability that the student has a phone.

   • Everything inside the first circle: 0.32 + 0.18 = 0.5

3. (c) Find the probability that the student has a tablet but no phone.

   • Only the second circle: 0.27

Two fair six-sided dice are rolled and the scores are added. Draw a sample space diagram and find the probability that

1. (a) the sum is 8

   • Draw the 6 by 6 grid of sums: 36 equally likely cells.
   • Cells with sum 8: 5
   • \(P=\frac{5}{36}\)

2. (b) the sum is less than 5

   • Cells with sum less than 5: 6
   • \(P=\frac{6}{36}=\frac{1}{6}\)

Two fair coins are tossed. Find the probability of

1. (a) exactly one head.

   • All outcomes: HH, HT, TH, TT
   • Favourable outcomes: 2
   • \(P=\frac{2}{4}=\frac{1}{2}\)

2. (b) at least one tail.

   • Outcomes with at least one T: 3
   • \(P=\frac{3}{4}\)

Three cards show the letters J, A and M. They are placed in a row in a random order. List all the arrangements and find the probability that the first letter is a consonant.

• All arrangements: JAM, JMA, AJM, AMJ, MJA, MAJ
• Favourable: JAM, JMA, MJA, MAJ (4)
• \(P=\frac{4}{6}=\frac{2}{3}\)

A bag holds 3 red counters and 5 blue counters. A counter is taken at random, its colour is noted and it is put back. A second counter is then taken. Find the probability that

1. (a) both counters are red.

   • With replacement the bag is the same for both draws: P(red) = 3/8 each time.
   • \(\frac38\times\frac38=\frac{9}{64}=0.141\)

2. (b) the two counters are different colours.

   • RB: \(\frac38\times\frac58=\frac{15}{64}\). BR: \(\frac58\times\frac38=\frac{15}{64}\)
   • Add: \(\frac{30}{64}=\frac{15}{32}=0.469\)

Machine X makes 60% of the bolts in a factory and machine Y makes the rest. 3% of the bolts from X are defective and 5% of the bolts from Y are defective. A bolt is chosen at random. Find the probability that it

1. (a) was made by X and is defective.

   • 0.6 × 0.03 = 0.018

2. (b) is defective.

   • Other path: 0.4 × 0.05 = 0.02
   • Add: 0.018 + 0.02 = 0.038

A box holds 5 milk, 4 dark and 3 white chocolates. Two chocolates are taken at random without replacement. Find the probability that both are the same type.

• Same type twice: three paths. After one is taken, 11 remain.
• \(P=\frac{5}{12}\times\frac{4}{11} + \frac{4}{12}\times\frac{3}{11} + \frac{3}{12}\times\frac{2}{11}\)
• \(P=\frac{19}{66}\)

An archer hits the target with each arrow with probability 0.8, independently. She shoots three arrows. Find the probability that she hits the target

1. (a) exactly twice.

   • Paths with two hits and one miss: HHM, HMH, MHH.
   • Each path: \(0.8\times0.8\times0.2=0.128\)
   • Add: \(3\times0.128=0.384\)

2. (b) at least once.

   • P(no hits) = \(0.2^3=0.008\)
   • \(1-0.008=0.992\)

A drawer holds 10 light bulbs, 3 of which are faulty. Two bulbs are taken without replacement. Find the probability that at least one is faulty.

• \(P(\text{none})=\frac{7}{10}\times\frac{6}{9}=\frac{7}{15}\)
• \(P(\text{at least one})=1-\frac{7}{15}=\frac{8}{15}\)

A fair four-sided die and a fair six-sided die are rolled and the scores are added. Find the probability that the sum is a prime number.

• Draw the 4 by 6 grid of sums: 24 cells.
• Cells where the sum is a prime number: 11
• \(P=\frac{11}{24}\)