Six cards, each with one letter, form the word "kareta". The…
Six cards, each with one letter, form the word "kareta". The cards are shuffled and then drawn one by one at random. What is the probability that the letters drawn in order form the word "raketa"?
Explanation of Method
The letters are k, a, r, e, t, a — the letter "a" appears twice. We compute the probability draw by draw with the multiplication rule: at each step, the probability that the required letter comes next is the number of remaining cards with that letter divided by the number of remaining cards. We check by counting orders of the six cards.
Step 1: The cards
k, a, r, e, t, a: six cards, with two cards "a" and one card of each other letter.
Step 2: Draw by draw
| Draw | Needed | Remaining cards | Probability |
|---|---|---|---|
| 1 | r | k a r e t a | $1/6$ |
| 2 | a | k a e t a | $2/5$ |
| 3 | k | k e t a | $1/4$ |
| 4 | e | e t a | $1/3$ |
| 5 | t | t a | $1/2$ |
| 6 | a | a | $1$ |
\[ \boxed{P=\frac16\cdot\frac25\cdot\frac14\cdot\frac13\cdot\frac12\cdot1=\frac{2}{720}=\frac1{360}\approx0.0028.} \]
Step 3: Check by counting
The $6$ cards can come out in $6!=720$ equally likely orders. The word "raketa" is obtained when the two "a" cards are in positions $2$ and $6$ (in either order) and the other cards in their places: $2$ orders. So $P=\frac2{720}=\frac1{360}$. ✓
Equivalently, there are $\frac{6!}{2!}=360$ different letter sequences, all equally likely, and "raketa" is one of them.
Result
| Answer | |
|---|---|
| Probability | $\frac16\cdot\frac25\cdot\frac14\cdot\frac13\cdot\frac12=\frac1{360}$ |