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Statistical probability · 3 marks

A tyre company records distances before replacement in 1000 cases. Find the probability that a randomly chosen tyre lasts:

  1. (i) Less than 4000 km.
  2. (ii) Between 4000 and 14000 km.
  3. (iii) More than 14000 km.
Distance(km)Less than40004001 to90009001 to14000More than14000Number ofcases20210325445
Answer: (i) 201000 = 0.02 (ii) 5351000 = 0.535 (iii) 4451000 = 0.445

Step-by-step solution

Given: Less than 4000 km: 20; 4001 to 9000 km: 210; 9001 to 14000 km: 325; more than 14000 km: 445; Total: 1000 cases

Idea: This is statistical (experimental) probability: P = Number of tyres in that groupTotal number of tyres. “Between 4000 and 14000 km” covers two columns of the table, so add them.

(i) Less than 4000 km.

  1. Tyres lasting less than 4000 km: 20. P = 201000 = 0.02.1 mark
0.02

(ii) Between 4000 and 14000 km.

  1. This is the 4001–9000 group and the 9001–14000 group together: 210 + 325 = 535.½ mark
  2. P = 5351000 = 0.535.½ mark
0.535

(iii) More than 14000 km.

  1. Tyres lasting more than 14000 km: 445. P = 4451000 = 0.445.1 mark
0.445
(i) 20/1000 = 0.02 (ii) 535/1000 = 0.535 (iii) 445/1000 = 0.445

Check: Total of the table: 20 + 210 + 325 + 445 = 1000 ✓. The three answers cover all the tyres: 0.02 + 0.535 + 0.445 = 1 ✓.

Answer to write in the exam

(i)

Total cases = 20 + 210 + 325 + 445 = 1000

∴ P(less than 4000 km) = 201000 = 0.02

(ii)

Number of tyres = 210 + 325 = 535

∴ P(between 4000 and 14000 km) = 5351000 = 0.535

(iii)

∴ P(more than 14000 km) = 4451000 = 0.445

Common mistakes that cost marks

  • In (ii), using only one column (0.21 or 0.325). “Between 4000 and 14000” includes both middle groups.
  • Dividing by 4 (the number of groups) instead of by 1000 (the number of tyres).
  • Writing 4451000 as 4.45 or 0.0445. Dividing by 1000 moves the decimal point three places: 0.445.

How this can come in the exam

Case-based (4 marks)

A bulb maker tests 500 bulbs and records how long they last: under 800 hours: 40; 800 to 1200 hours: 180; over 1200 hours: 280.
(a) Find the probability that a bulb lasts under 800 hours. (1)
(b) Find the probability that a bulb lasts over 1200 hours. (1)
(c) A shop buys 2000 such bulbs. Estimate how many will last at least 800 hours. (2)

Show answer(a) 40500 = 0.08.
(b) 280500 = 0.56.
(c) P(at least 800 h) = (180 + 280)/500 = 460500 = 0.92; estimate = 0.92 × 2000 = 1840 bulbs.
MCQ (1 mark)

Out of 400 phones checked, 18 had a faulty screen. The probability that a phone chosen at random has a faulty screen is

  1. 0.018
  2. 0.045
  3. 0.18
  4. 0.45
Show answer

(B) 0.045
18400 = 0.045.

Try one yourself

In 200 cases, a battery lasted under 1 year in 30 cases, 1 to 2 years in 110 cases and over 2 years in 60 cases. Find the probability that a battery lasts at least 1 year.

Show answer

(110 + 60)/200 = 170200 = 0.85.

More questions like this

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