1988 AMC 12 Problem 28

Question

An unfair coin has probability

p

of coming up heads on a single toss. Let

w

be the probability that, in

5

independent tosses of this coin, heads come up exactly

3

times. If

w=\frac{144}{625},

then

p

must be

\frac{2}{5}

p

must be

\frac{3}{5}

p

must be greater than

\frac{3}{5}

p

is not uniquely determined

there is no value of

p

for which

w=\frac{144}{625}

Write

w

as a function of

p

using the binomial coefficient

Check one simple value, then compare the function at

p=\frac{3}{5}

and

p=1

Here

w(p)=10p^3(1-p)^2.

At

p=\frac{2}{5},

w=10\left(\frac25\right)^3\left(\frac35\right)^2 =\frac{144}{625}.

Also

w(\frac{3}{5})=\frac{216}{625}>\frac{144}{625},

while

w(1)=0.

By continuity there is another solution between

\frac{3}{5}

and

1.

Hence

p

is not unique.

Thus the correct answer is D.

Problem 28 in other years