Porisma

Worked solution

Condition a hypergeometric count on at least one match

A box contains 45 balls distinctly labelled 1 to 45. Seven are drawn randomly without replacement and their labels are the winning numbers. Andy randomly chooses eight distinct integers from 1 to 45.

(i) Show that the probability none of Andy’s numbers matches the winning numbers is 0.227. mark

(ii) Hence find the probability exactly four match, given at least one match. marks

[2014/MJC/II/5]

CJC · Review problems · Question 9

Answer

(i) 0.227; (ii) 0.0155 (both 3 s.f.).

(i) Choose all eight from the 38 non-winning labels

For any fixed winning set, all (458)\binom{45}8 possible Andy selections are equally likely. None matches in (388)\binom{38}8 selections. Thus P(N)=(388)/(458)≈0.226874≈0.227P(N)=\binom{38}8/\binom{45}8\approx0.226874\approx0.227 (3 s.f.).

How did this part go?

(ii) Count exactly four winning and four non-winning labels

P(four matches)=(74)(384)/(458)P(\text{four matches})=\binom74\binom{38}4/\binom{45}8. Four matches already implies at least one, so divide by 1−P(N)1-P(N):

P(four∣at least one)=(74)(384)/(458)1−(388)/(458)≈0.015502≈0.0155(3 s.f.).P(\text{four}\mid\text{at least one})=\frac{\binom74\binom{38}4/\binom{45}8}{1-\binom{38}8/\binom{45}8}\approx0.015502\approx0.0155\quad(3\text{ s.f.}).

Keep the unrounded ratio from (i), rather than substituting its 3 s.f.answer 0.227.

How did this part go?

Common mistakes

Andy chooses eight labels although there are seven winning labels; use the correct size in the common denominator.

Solution notes

Both asks, full combinations/conditional denominator, source intermediate decimals, final roundings and marks retained; exact forms used to avoid premature rounding.

Read the hint