Probability from two-way tables Problem set B Worked solutions

The table shows the members of a fitness centre who joined each class in one week. One member is chosen at random. Find the probability that:

1. (a) the member joined swim;

   • Swim row: 28 + 14 = 42
   • Grand total: 53 + 55 + 42 = 150
   • \(P=\frac{42}{150}=\frac{7}{25}=0.28\)

2. (b) the member came in the morning.

   • Morning column: 22 + 17 + 28 = 67
   • Grand total: 150
   • \(P=\frac{67}{150}\)

The table shows the drink and size ordered by customers at a café in one hour. One customer is chosen at random. Find the probability that the customer:

1. (a) ordered coffee and ordered a medium drink;

   • Both properties is one cell: Coffee and Medium = 48
   • Grand total: 175
   • \(P=\frac{48}{175}\)

2. (b) ordered tea or ordered a large drink (or both).

   • Tea row total 75, Large column total 45; the cell they share (15) is counted twice
   • Number: 75 + 45 − 15 = 105
   • \(P=\frac{105}{175}=\frac{3}{5}=0.6\)

The table shows the type of pet and the type of home of the families in a survey.

1. (a) One family who has a dog is chosen at random. Find the probability that this family lives in a flat.

   • Only the Dog row is left: total 9 + 41 = 50
   • Of these, Flat: 9
   • \(P=\frac{9}{50}=0.18\)

2. (b) One family who lives in a house is chosen at random. Find the probability that this family has fish.

   • Only the House column is left: total 26 + 41 + 9 = 76
   • Of these, Fish: 9
   • \(P=\frac{9}{76}\)

The table shows the flavour and the size of the ice creams sold at a kiosk. One ice cream that is in a tub is chosen at random. Find the probability that this ice cream is vanilla.

• Only the Tub column is left: 16 + 32 = 48
• Of these, Vanilla: 16
• \(P=\frac{16}{48}=\frac{1}{3}\)

The table shows the juice bottles on a shop shelf, by flavour and size. There are 150 bottles in all, and the number in the Orange row, Large column is k.

1. (a) Find the value of k.

   • Known cells total 106, so k = 150 − 106 = 44

2. (b) One bottle is chosen at random. Find the probability that the bottle is orange.

   • Orange row: 44 + 30 = 74
   • \(P=\frac{74}{150}=\frac{37}{75}\)

3. (c) One bottle that is small is chosen at random. Find the probability that this bottle is berry.

   • Small column total: 68
   • Of these, Berry: 22
   • \(P=\frac{22}{68}=\frac{11}{34}\)

The table shows the sport played and the age group of the players in a league. One player is chosen at random.

1. (a) Find the probability that the player does not play football.

   • Football row total: 55; so the others: 150 − 55 = 95
   • \(P=\frac{95}{150}=\frac{19}{30}\)

2. (b) Given that the player does not play football, find the probability that the player is 16 or over.

   • Keep the other rows (Rugby, Hockey): total 95
   • Of these, 16 or over: 26 + 29 = 55
   • \(P=\frac{55}{95}=\frac{11}{19}\)

One hundred and sixty customers answered a survey about a new app. The table is partly filled in.

1. (a) Copy the table and find the missing numbers.

   • Liked it row: 96 − 70 = 26
   • Under 30 column: 100 − 70 = 30
   • Did not like it row: 64 − 30 = 34

2. (b) One customer is chosen at random. Find the probability that the customer liked the app and is 30 or over.

   • One cell: 26; grand total 160
   • \(P=\frac{26}{160}=\frac{13}{80}=0.1625\)

3. (c) One customer who is under 30 is chosen at random. Find the probability that this customer did not like the app.

   • Under 30 column total: 100
   • Of these, Did not like it: 30
   • \(P=\frac{30}{100}=\frac{3}{10}=0.3\)