A card is drawn from a bag of cards numbered 1, 2, 3, recorded, and returned; this is repeated five times. The five recorded numbers are sorted in decreasing order as a >= b >= c >= d >= e. Find the probability that b = 3.

Step-by-step solution

1. 1

   Translate the order statistic into a count. Sorting in decreasing order makes $b$ the second largest value. Since $3$ is the largest possible card, $b=3$ happens exactly when at least two of the five draws show a $3$. (If only one $3$ appeared, $a$ would be $3$ but $b$ would be $2$ or $1$.)

   $\{b=3\}=\{X\ge 2\},\qquad X=\#\text{ of }3\text{s}$

2. 2

   Identify the distribution of X. Draws are with replacement, so $X\sim\text{Binomial}(n=5,\ p=\tfrac13)$.

3. 3

   Use the complement — two terms instead of four. Rather than adding $P(X=2)+P(X=3)+P(X=4)+P(X=5)$, subtract the two small cases:

   $P(X\ge 2)=1-P(X=0)-P(X=1)$

4. 4

   Compute the two subtracted terms.

   $P(X=0)=\left(\frac23\right)^{5}=\frac{32}{243},\qquad P(X=1)=\binom51\left(\frac13\right)\left(\frac23\right)^{4}=\frac{80}{243}$

5. 5

   Subtract.

   $P(b=3)=1-\frac{32}{243}-\frac{80}{243}=\frac{243-112}{243}=\frac{131}{243}\approx 0.539$

   $\boxed{\dfrac{131}{243}}$

6. 6

   Check against the companion case b = 1. By the mirror argument, $b=1$ requires at most one draw to be $2$ or $3$, giving $\left(\tfrac13\right)^5+5\left(\tfrac23\right)\left(\tfrac13\right)^4=\tfrac{1+10}{243}=\tfrac{11}{243}$. Since $b$ must be $1$, $2$ or $3$, the remaining case is $P(b=2)=1-\tfrac{131}{243}-\tfrac{11}{243}=\tfrac{101}{243}$, and all three sum to $1$ ✓.