Challenge question #55 (17/07/2017)
Junior Question
Four students have forgotten to write their names on their maths test. Their teacher hands out the marked tests to the four students at random.
What is the probability that none of the students receive the correct test?
$\tfrac{3}{8}$
Worked SolutionLet $A$, $B$, $C$, and $D$ represent the four students. There are $4! = 24$ ways of ordering $A$, $B$, $C$, and $D$:
A student receives the correct test if they are in the same position as the order of their letter in the alphabet. For example, if $A$ is in the first position, or if $B$ is in the second position, and so on. We remove the orderings where at least one letter is in its correct position.
There are $9$ outcomes where no letter is in its correct position.
So, the probability that no student receives the correct test is $\tfrac{9}{24} = \tfrac{3}{8}$.
Senior Question
Jasba and Chris play a game with an unusual set of dice.
- Die $A$ has sides $2$, $2$, $4$, $4$, $9$, $9$
- Die $B$ has sides $1$, $1$, $6$, $6$, $8$, $8$
- Die $C$ has sides $3$, $3$, $5$, $5$, $7$, $7$
Jasba chooses his die first, then Chris chooses his die from those remaining. Both players roll their die, and the player who rolls the higher number wins.
Describe the strategy Chris should use to ensure that he always has an advantage in the game.- Die $A$, Chris should choose die $C$
- Die $B$, Chris should choose die $A$
- Die $C$, Chris should choose die $B$
We consider all of the possible outcomes that can occur when two of the dice are rolled against each other.
- Die $A$, Chris should choose die $C$
- Die $B$, Chris should choose die $A$
- Die $C$, Chris should choose die $B$
In each case, the probability that Chris wins the game is $\tfrac{5}{9}$, while Jasba has probability $\tfrac{4}{9}$.
