Mixed practice Problem set A One question at a time

The table shows the favourite drink of adults and children at a fair.

1. (a) Find the probability that a person chosen at random prefers juice.

   • Total people: 150. Juice column: 18 + 44 = 62.
   • \(P=\frac{62}{150}=\frac{31}{75}\)

2. (b) Find the probability that a person chosen at random is an adult.

   • Adult row: 31 + 18 + 22 = 71.
   • \(P=\frac{71}{150}\)

3. (c) Find the probability that a person chosen at random is an adult who prefers juice.

   • The adult-and-juice cell is 18.
   • \(P=\frac{18}{150}=\frac{3}{25}\)

Events A and B are such that

1. (a) \(P(A)=0.55\), \(P(A\cap B)=0.15\) and \(P(A\cup B)=0.8\). Find \(P(B)\).

   • \(P(B)=P(A\cup B)-P(A)+P(A\cap B)\)
   • \(=0.8-0.55+0.15=0.4\)

2. (b) \(P(A)=0.5\), \(P(A\cup B)=0.85\) and \(P(B')=0.6\). Find \(P(A\cap B)\).

   • \(P(B)=1-0.6=0.4\)
   • \(P(A\cap B)=0.5+0.4-0.85=0.05\)

A drawer holds 8 black socks and 6 white socks. Two socks are taken out at random without replacement. Find the probability that they are different colours.

• Total 14. Different colours: black then white, or white then black.
• \(P=\frac{8}{14}\times\frac{6}{13}+\frac{6}{14}\times\frac{8}{13}\)
• \(=\frac{48}{91}\)

A gym records when members train and what they do.

1. (a) Find the probability that a member uses weights, given that the member trains in the morning.

   • \(\frac{18}{60}=\frac{3}{10}\)

2. (b) Find the probability that a member uses weights, given that the member trains in the evening.

   • \(\frac{12}{40}=\frac{3}{10}\)

3. (c) Determine whether using weights is independent of the time of day. Justify your answer.

   • If independent, the two conditional probabilities equal each other and equal \(P(\text{weights})=\frac{30}{100}\).
   • \(\frac{3}{10}\)
   • They are equal, so the events are independent.

Of 36 students, 21 study physics, 8 study physics and chemistry, and 9 study neither subject.

1. (a) Find the number of students that study chemistry but do not study physics.

   • Draw the Venn diagram: 8 in both, 21 in physics, 9 in neither.
   • physics only: 21 − 8 = 13
   • chemistry only: 36 − 21 − 9 = 6

2. (b) Find the probability that a student chosen at random studies chemistry.

   • Everyone in the chemistry circle: 6 + 8 = 14
   • \(P=\frac{14}{36}=\frac{7}{18}\)

3. (c) Given that a student studies chemistry, find the probability that they study physics.

   • Restrict to the chemistry circle: 14 people.
   • \(P=\frac{8}{14}=\frac{4}{7}\)

Salma goes to school by bus with probability 0.65 and otherwise walks. She is late with probability 0.15 if she takes the bus and 0.05 if she walks.

1. (a) Find the probability that Salma takes the bus and is late.

   • Multiply along the branch: \(0.65\times 0.15=0.0975\)

2. (b) Find the probability that Salma is on time.

   • \(P(\text{late})=0.65\times 0.15+0.35\times 0.05=0.0975+0.0175=0.115\)
   • \(P(\text{on time})=1-0.115=0.885\)

A café has three chefs. Each customer is served by one of them, each with probability \(\frac{1}{3}\). The probability that the meal arrives cold is 0.05 for Ana, 0.12 for Bo and 0.2 for Cy.

1. (a) Find the probability that a customer is served by Cy and the meal is cold.

   • \(\frac{1}{3}\times 0.2=\frac{1}{15}\)

2. (b) Find the probability that a customer's meal arrives hot.

   • \(P(\text{cold})=\frac{1}{3}(0.05+0.12+0.2)=\frac{37}{300}\)
   • \(P(\text{hot})=1-\frac{37}{300}=\frac{263}{300}\)

In a population, 2% of people have a certain disease. A test gives a positive result for 95% of people who have the disease and for 8% of people who do not. A person is chosen at random and tested.

1. (a) Find the probability that the test is positive.

   • Disease and positive: \(0.02\times0.95=0.019\)
   • No disease and positive: \(0.98\times0.08=0.0784\)
   • Add: \(0.019+0.0784=0.0974\)

2. (b) Given that the test is positive, find the probability that the person has the disease.

   • \(P(\text{disease}\mid\text{positive})=\dfrac{0.019}{0.0974}=0.195\)

Three fair six-sided dice are thrown and the scores are added. Find the probability that the sum is 6.

• List the sets of three scores that add to 6, and count the orders of each.
• 1 + 1 + 4: 3 ways; 1 + 2 + 3: 6 ways; 2 + 2 + 2: 1 way
• Total 10 outcomes out of \(6^3=216\)
• \(P=\frac{10}{216}=\frac{5}{108}\)

Events A and B are such that \(P(A)=\frac{2}{3}\), \(P(B|A)=\frac{1}{4}\) and \(P(A\cup B)=\frac{5}{6}\).

1. (a) Find \(P(A\cap B)\).

   • \(P(A\cap B)=P(A)\times P(B|A)=\frac{2}{3}\times \frac{1}{4}=\frac{1}{6}\)

2. (b) Find \(P(B)\).

   • \(P(B)=P(A\cup B)-P(A)+P(A\cap B)\)
   • \(=\frac{5}{6}-\frac{2}{3}+\frac{1}{6}=\frac{1}{3}\)

The probability that it rains on Saturday is 0.3. If it rains on Saturday, the probability that it rains on Sunday is 0.6. If it is dry on Saturday, the probability that it rains on Sunday is 0.2.

1. (a) Find the probability that it rains on Sunday.

   • Rain both days: \(0.3\times0.6=0.18\). Dry Saturday, rain Sunday: \(0.7\times0.2=0.14\)
   • Add: \(0.18+0.14=0.32\)

2. (b) Given that it rains on Sunday, find the probability that it rained on Saturday.

   • \(P(\text{rain Sat}\mid\text{rain Sun})=\dfrac{0.18}{0.32}=0.5625=0.563\)