Question
A festival prints promotional tickets using one of two machines, selected at random afresh for each ticket. Machine A prints Sponsor Gold's mark on 80% of its tickets and Sponsor Blue's mark on 20%. Machine B prints Gold's mark on 20% and Blue's on 60%; its other tickets are blank. Each ticket carries at most one sponsor mark.
The organiser wants exactly 50% of all tickets to show Gold and exactly 50% to show Blue in the long-run probability sense. The only control is the single probability \(q\) of choosing machine A, with \(1-q\) for machine B; the machines' internal behaviour cannot be changed.
Can one valid value of \(q\) meet both promises? If not, calculate what each promise separately demands and explain the incompatibility.
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Hint 1
A ticket’s chance of carrying each mark is a weighted average of the two machines’ chances, using the same mixing weight.
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Solution
The organiser has only one decision variable, \(q=P(\text{use machine A})\), shared by both sponsor outcomes. By conditioning on the selected machine,
\[
P(G)=0.8q+0.2(1-q)=0.2+0.6q,
\qquad P(B)=0.2q+0.6(1-q)=0.6-0.4q.
\]
To promise a 50% Gold rate, \(0.2+0.6q=0.5\) requires \(q=0.3/0.6=\boxed{1/2}\). But the 50% Blue rate requires \(0.6-0.4q=0.5\), or \(q=0.1/0.4=\boxed{1/4}\).
Both weights are individually feasible, but they differ. A machine-selection policy cannot simultaneously use machine A with probability 1/2 and 1/4. If the organiser honours Gold, Blue's rate would be \(0.6-0.4(1/2)=0.4\), only 40%. If they honour Blue, Gold's rate would be \(0.2+0.6(1/4)=0.35\), only 35%.
At the permitted endpoints \(q=0\) and \(q=1\), the output pairs are \((\text{Gold }20\%,\text{ Blue }60\%)\) and \((\text{Gold }80\%,\text{ Blue }20\%)\). Every achievable pair lies on the line segment connecting those points; \((50\%,50\%)\) is not on that segment. Thus the promises are incompatible under the one-parameter policy, even though each separate target is attainable.
Assumptions
The two machines’ stated conditional ticket-mark probabilities are fixed.
Every ticket uses exactly one machine, selected independently with the same probability q.
A ticket carries at most one mark, with the remaining machine-B probability producing a blank ticket.
Final Answer
No. Gold at 50% requires \(q=\frac12\), while Blue at 50% requires \(q=\frac14\). One selection probability cannot satisfy both.