1992 AMC 12 Problem 29

Question

An “unfair” coin has a

\frac23

probability of turning up heads. If this coin is tossed

50

times, what is the probability that the total number of heads is even?

2^5\left(\frac23\right)^{50}

\frac12\left(1-\frac1{3^{50}}\right)

\frac12

\frac12\left(1+\frac1{3^{50}}\right)

\frac23

Track the difference between the probabilities of an even and an odd number of heads

One toss multiplies that difference by

P(T)-P(H)

Let

E_n

and

O_n

be the probabilities of an even and odd number of heads after

n

tosses. Then

E_n+O_n=1,

while

\begin{aligned} E_{n+1}-O_{n+1} &=\left(\frac13-\frac23\right)\\ &\qquad\cdot(E_n-O_n). \end{aligned}

Since

E_0-O_0=1,

we have

E_{50}-O_{50}=(-\frac{1}{3})^{50}=3^{-50}.

Solving the sum and difference equations gives

E_{50}=\frac12\left(1+\frac1{3^{50}}\right).

Thus the correct answer is D.

Problem 29 in other years