When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to $0$ and is the same as that of getting heads exactly twice. Let $\frac ij$, in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j$.

Denote the probability of getting a heads in one flip of the biased coin as $h$. Based upon the problem, note that ${5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3$. After canceling out terms, we get $1 - h = 2h$, so $h = \frac{1}{3}$. The answer we are looking for is ${5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}$, so $i+j=40+243=\boxed{283}$.
283
Level 5
Counting & Probability
Problem: When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to $0$ and is the same as that of getting heads exactly twice. Let $\frac ij$, in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j$.

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>Let the probability of getting heads in a single flip be denoted as \( p \). The probability of getting tails is then \( 1 - p \). The number of ways to get exactly \( k \) heads in \( n \) flips is given by the binomial coefficient \( \binom{n}{k} \). 

For \( n = 5 \), the probability of getting exactly \( k \) heads is given by the formula:

\[
P(X = k) = \binom{5}{k} p^k (1 - p)^{5 - k}
\]

We need to find the probabilities for \( k = 1 \) and \( k = 2 \):

1. For \( k = 1 \):
\[
P(X = 1) = \binom{5}{1} p^1 (1 - p)^{4} = 5p(1 - p)^{4}
\]

2. For \( k = 2 \):
\[
P(X = 2) = \binom{5}{2} p^2 (1 - p)^{3} = 10p^2(1 - p)^{3}
\]

According to the problem, these two probabilities are equal:

\[
5p(1 - p)^{4} = 10p^2(1 - p)^{3}
\]

We can simplify this equation. First, divide both sides by \( p \) (noting \( p \neq 0 \)):

\[
5(1 - p)^{4} = 10p(1 - p)^{3}
\]

Next, divide both sides by \( (1 - p)^{3} \) (noting \( 1 - p \neq 0 \)):

\[
5(1 - p) = 10p
\]

Rearranging gives:

\[
5 - 5p = 10p
\]
\[
5 = 15p
\]
\[
p = \frac{1}{3}
\]

Now, we need to find the probability of getting exactly 3 heads:

\[
P(X = 3) = \binom{5}{3} p^3 (1 - p)^{2} = 10p^3(1 - p)^{2}
\]

Substituting \( p = \frac{1}{3} \):

\[
P(X = 3) = 10 \left(\frac{1}{3}\right)^3 \left(1 - \frac{1}{3}\right)^{2} = 10 \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^{2}
\]

Calculating \( P(X = 3) \):

\[
= 10 \cdot \frac{1}{27} \cdot \frac{4}{9} = 10 \cdot \frac{4}{243} = \frac{40}{243}
\]

The fraction \( \frac{40}{243} \) is already in lowest terms. Thus, \( i = 40 \) and \( j = 243 \). 

Finally, we compute \( i + j \):

\[
i + j = 40 + 243 = 283
\]<|END_OF_LATENT|> The final answer is $283$.