UCAT independent and dependent events

Independent: multiply the probabilities. Without replacement: after the first pick, take one away from the top and the bottom. At least one: 1 minus the chance of none. Try two below.

By Shrey, Cambridge studentUpdated
Decision Making · Question 1 of 20:00 / 1:03

A bag contains 8 red and 10 blue counters. Two counters are selected at random without replacement. What is the probability that both are red, to one decimal place?

Worked answers

1. A bag contains 8 red and 10 blue counters. Two counters are selected at random without replacement. What is the probability that both are red, to one decimal place?

Answer: B: 18.3%

  1. First red probability = 8/18.
  2. Second red after a red = 7/17.
  3. Both red = 8/18 × 7/17 = 18.3%.
2. Three independent checks each have a 18% chance of identifying a fault. What is the probability that at least one identifies it, to one decimal place?

Answer: B: 44.9%

  1. No check identifies it = 0.82³.
  2. At least one identifies it = 1 − 0.551368 = 44.9%.

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Three rules

SituationDo thisExample
Both happen, independentMultiply0.6 × 0.6 = 0.36
Without replacementChange the second fraction8/18 × 7/17
At least one1 − P(none)1 − 0.82³ = 0.449

The trap

Adding probabilities for "at least one" goes wrong: three 40% chances do not make 120%. Work out none, then subtract from 1.

Keep going

Given that questions: UCAT conditional probability

Every Decision Making type: DM question types and methods

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