Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 16 / a

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

i

Solution

Random sampling without replacement assigns each of the ten distinguishable pieces to each lunch with the same probability. Five pieces of each type therefore give

P (A_{1}) = \frac{1}{2}, P (B_{2}) = \frac{1}{2} .

One can check the second probability by conditioning on the first lunch:

P (B_{2}) = P (B_{2} ∣ A_{1}) P (A_{1}) + P (B_{2} ∣ B_{1}) P (B_{1}) = \frac{5}{9} \frac{1}{2} + \frac{4}{9} \frac{1}{2} = \frac{1}{2} .

The identical marginal probabilities do not imply independence of the draws.

ii

Solution

Condition on an apple occupying position two in a uniformly random ordering. The nine other positions then contain four apples and five bananas, each equally likely to occupy position three. The conditional probability is therefore

P (A_{3} ∣ A_{2}) = \frac{4}{9} .

This uses exchangeability of the unconditioned positions; there is no need to assume the first lunch contained a particular fruit.

iii

Solution

Each position has apple probability

1/2

. Applying Bayes' theorem to the preceding conditional probability gives

P (A_{2} ∣ A_{3}) = \frac{P ( A _{3} ∣ A _{2} ) P ( A _{2} )}{P ( A _{3} )} = \frac{( 4/9 ) ( 1/2 )}{1/2} = \frac{4}{9} .

This also agrees with exchangeability of the second and third positions.

iv

Solution

By the inclusion-exclusion principle for probabilities,

P (B_{1} \cup A_{2}) = P (B_{1}) + P (A_{2}) - P (B_{1} \cap A_{2}) .

After an initial banana, five of the nine remaining pieces are apples, so the intersection probability is

(1/2) (5/9) = 5/18

. Consequently

P (B_{1} \cup A_{2}) = \frac{1}{2} + \frac{1}{2} - \frac{5}{18} = \frac{13}{18} .

As a check, the complementary event is an apple followed by a banana, which also has probability

5/18

.

v

Solution

Let

I_{j}

be the indicator random variable of an apple in lunch

j

. The number remaining is

R = 5 - I_{1} - I_{2} - I_{3}

. Each expected value

E [I_{j}] = P (A_{j}) = 1/2

. The linearity of expectation therefore gives

E [R] = 5 - 3 (\frac{1}{2}) = \frac{7}{2} apples .

Although the indicator random variables are dependent under sampling without replacement, the linearity of expectation needs no independence assumption.

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