Mixed practice Problem set B One question at a time

The table shows the fuel type and colour of cars in a car park.

1. (a) Find the probability that a car chosen at random is silver.

   • Total cars: 120. Silver column: 26 + 21 = 47.
   • \(P=\frac{47}{120}\)

2. (b) Find the probability that a car chosen at random is a hybrid.

   • Hybrid row: 14 + 26 + 9 = 49.
   • \(P=\frac{49}{120}\)

3. (c) Find the probability that a car chosen at random is a silver hybrid.

   • The hybrid-and-silver cell is 26.
   • \(P=\frac{26}{120}=\frac{13}{60}\)

Events A and B are such that

1. (a) \(P(A)=0.45\), \(P(A\cap B)=0.2\) and \(P(A\cup B)=0.75\). Find \(P(B)\).

   • \(P(B)=P(A\cup B)-P(A)+P(A\cap B)\)
   • \(=0.75-0.45+0.2=0.5\)

2. (b) \(P(A)=0.5\), \(P(A\cup B)=0.85\) and \(P(B')=0.55\). Find \(P(A\cap B)\).

   • \(P(B)=1-0.55=0.45\)
   • \(P(A\cap B)=0.5+0.45-0.85=0.1\)

A bag holds 10 red sweets and 5 green sweets. Two sweets are taken out at random without replacement. Find the probability that they are different colours.

• Total 15. Different colours: red then green, or green then red.
• \(P=\frac{10}{15}\times\frac{5}{14}+\frac{5}{15}\times\frac{10}{14}\)
• \(=\frac{10}{21}\)

A sixth-form common room records the hot drink chosen by students.

1. (a) Find the probability that a student chooses tea, given that the student is in Year 12.

   • \(\frac{24}{40}=\frac{3}{5}\)

2. (b) Find the probability that a student chooses tea, given that the student is in Year 13.

   • \(\frac{15}{40}=\frac{3}{8}\)

3. (c) Determine whether choosing tea is independent of year group. Justify your answer.

   • If independent, the two conditional probabilities equal each other and equal \(P(\text{tea})=\frac{39}{80}\).
   • \(\frac{3}{5}\ne \frac{3}{8}\)
   • They are not equal, so the events are not independent.

Of 48 houses on an estate, 26 have solar panels, 11 have solar panels and a heat pump, and 10 have neither.

1. (a) Find the number of houses that have a heat pump but do not have solar panels.

   • Draw the Venn diagram: 11 in both, 26 in solar panels, 10 in neither.
   • solar panels only: 26 − 11 = 15
   • a heat pump only: 48 − 26 − 10 = 12

2. (b) Find the probability that a house chosen at random has a heat pump.

   • Everyone in the heat pump circle: 12 + 11 = 23
   • \(P=\frac{23}{48}\)

3. (c) Given that a house has a heat pump, find the probability that they have solar panels.

   • Restrict to the heat pump circle: 23 people.
   • \(P=\frac{11}{23}\)

A courier delivers a parcel by van with probability 0.7 and otherwise by bike. The parcel arrives damaged with probability 0.04 by van and 0.01 by bike.

1. (a) Find the probability that the parcel goes by van and arrives damaged.

   • Multiply along the branch: \(0.7\times 0.04=0.028\)

2. (b) Find the probability that the parcel arrives intact.

   • \(P(\text{damaged})=0.7\times 0.04+0.3\times 0.01=0.028+0.003=0.031\)
   • \(P(\text{intact})=1-0.031=0.969\)

A quiz team has three members. Each question is answered by one of them, each with probability \(\frac{1}{3}\). The probability that the answer is wrong is 0.08 for Kai, 0.15 for Lu and 0.25 for Max.

1. (a) Find the probability that a question is answered by Max and the answer is wrong.

   • \(\frac{1}{3}\times 0.25=\frac{1}{12}\)

2. (b) Find the probability that a question is answered correctly.

   • \(P(\text{wrong})=\frac{1}{3}(0.08+0.15+0.25)=\frac{4}{25}\)
   • \(P(\text{right})=1-\frac{4}{25}=\frac{21}{25}\)

A factory makes phones and 3% of them are faulty. An automatic check flags 90% of the faulty phones and 6% of the phones that are not faulty. A phone is chosen at random and checked.

1. (a) Find the probability that the phone is flagged.

   • Faulty and flagged: \(0.03\times0.9=0.027\)
   • Not faulty and flagged: \(0.97\times0.06=0.0582\)
   • Add: \(0.027+0.0582=0.0852\)

2. (b) Given that the phone is flagged, find the probability that it is faulty.

   • \(P(\text{faulty}\mid\text{flagged})=\dfrac{0.027}{0.0852}=0.317\)

Three fair six-sided dice are rolled and the scores are added. Find the probability that the sum is 7.

• List the sets of three scores that add to 7, and count the orders of each.
• 1 + 1 + 5: 3 ways; 1 + 2 + 4: 6 ways; 1 + 3 + 3: 3 ways; 2 + 2 + 3: 3 ways
• Total 15 outcomes out of \(6^3=216\)
• \(P=\frac{15}{216}=\frac{5}{72}\)

Events A and B are such that \(P(A)=\frac{3}{4}\), \(P(B|A)=\frac{2}{3}\) and \(P(A\cup B)=\frac{7}{8}\).

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

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

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

   • \(P(B)=P(A\cup B)-P(A)+P(A\cap B)\)
   • \(=\frac{7}{8}-\frac{3}{4}+\frac{1}{2}=\frac{5}{8}\)

The probability that a train is late on Monday is 0.25. If it is late on Monday, the probability that it is late on Tuesday is 0.7. If it is on time on Monday, the probability that it is late on Tuesday is 0.1.

1. (a) Find the probability that the train is late on Tuesday.

   • Late both days: \(0.25\times0.7=0.175\). On time Monday, late Tuesday: \(0.75\times0.1=0.075\)
   • Add: \(0.175+0.075=0.25\)

2. (b) Given that the train is late on Tuesday, find the probability that it was late on Monday.

   • \(P(\text{late Mon}\mid\text{late Tue})=\dfrac{0.175}{0.25}=0.7\)