Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 11 / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii
- iv b
  - Solution iv
- v b
  - Solution v

i

Solution

There are two red counters among seven equally likely first draws, so

P (E_{1}) = \frac{2}{7} .

ii

Solution

Use the law of total probability according to the first draw. Without replacement, the second-draw red probability is

1/6

after a red and

2/6

after a green. Thus

P (E_{2}) = \frac{2}{7} \frac{1}{6} + \frac{5}{7} \frac{2}{6} = \frac{2}{7} .

Equivalently, each counter has the same probability of occupying the second position as the first. Equal marginal probabilities do not make the two draw events independent.

iii

Solution

After a red first draw, one red remains among six counters. The multiplication rule for conditional probability gives

P (E_{1} \cap E_{2}) = P (E_{1}) P (E_{2} ∣ E_{1}) = \frac{2}{7} \frac{1}{6} = \frac{1}{21} .

iv

Solution

By inclusion-exclusion, subtract the intersection once to correct the double counting:

P (E_{1} \cup E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2}) = \frac{2}{7} + \frac{2}{7} - \frac{1}{21} = \frac{11}{21} .

As a check, the complementary event is two greens, of probability

(5/7) (4/6) = 10/21

.

v

Solution

The definition of conditional probability gives

P (E_{1} ∣ E_{2}) = \frac{P ( E _{1} \cap E _{2} )}{P ( E _{2} )} = \frac{1/21}{2/7} = \frac{1}{6} .

Conditioning on a red in the second position leaves one red among the six counters that could occupy the first position, giving the same result directly.

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