A card is drawn from a bag containing cards numbered 1, 2 and 3, its number is recorded, and the card is returned. This is repeated five times. Find the probability that a 3 is recorded exactly once.

Step-by-step solution

1. 1

   Check that the trials are binomial. The card is replaced each time, so the five draws are independent and each has the same probability $p=\tfrac13$ of showing a $3$. With a fixed number of trials $n=5$ and a count of successes, the binomial model applies:

   $P(X=k)=\binom{n}{k}p^{k}(1-p)^{\,n-k}$

2. 2

   Choose which trial produces the 3. The single $3$ can occur on any one of the five draws:

   $\binom{5}{1}=5$

   Omitting this factor is the most common error and would give an answer five times too small.

3. 3

   Multiply the probabilities for one fixed pattern. For a specific choice of which draw is the $3$: that draw has probability $\tfrac13$, and each of the other four must show $1$ or $2$, probability $\tfrac23$ each:

   $\frac13\left(\frac23\right)^{4}=\frac13\cdot\frac{16}{81}=\frac{16}{243}$

4. 4

   Combine.

   $P=5\cdot\frac{16}{243}=\frac{80}{243}\approx 0.329$

   $\boxed{\dfrac{80}{243}}$

5. 5

   Check against the whole binomial distribution. The five probabilities $\binom5k(\tfrac13)^k(\tfrac23)^{5-k}$ for $k=0,\ldots,5$ are $\tfrac{32}{243},\tfrac{80}{243},\tfrac{80}{243},\tfrac{40}{243},\tfrac{10}{243},\tfrac{1}{243}$, and they sum to $\tfrac{243}{243}=1$ ✓. The expected number of $3$s is $np=\tfrac53\approx 1.67$, so $k=1$ being one of the two most likely outcomes is exactly what one would expect.