Problem B and Solution

A Pocketful of Coins

Dakarai has a some Canadian coins in his pocket: one nickel (worth \(\$0.05\)), one dime (worth \(\$0.10\)), one quarter (worth \(\$0.25\)), one loonie (worth \(\$1.00\)), and one toonie (worth \(\$2.00\)).

Suppose he reaches into his pocket and pulls out one coin at random.

What is the probability that he will pull out

a nickel?

a quarter?

a toonie?

What is the probability that the total value of the coins remaining in his pocket is

less than \(\$1.00\)?

greater than \(\$1.35\)?

less than \(\$2.00\)?

Suppose Dakarai reaches into his pocket and pulls out two coins at random. Which is greater, the probability that the coins in his hand have a value of \(\$0.35\), or the probability that the coins in his hand have a value of \(\$3.00\)?

Dakarai is selecting one of the five coins `at random’.

Since his selection is `at random’, there is an equal chance he will pull out any one of the coins, so the probability for each of these is equal to \(\frac{1}{5}=0.2\), or \(20\%\).

There is no combination of any four of the coins that has a total value less than \(\$1.00\). Therefore, this probability is equal to \(0\).

If Dakarai draws the coin of greatest value (the toonie), the total value of the remaining coins will be \(\$(1.00+0.25+0.10+0.05)=\$1.40\), which is greater than \(\$1.35\). So the total value of the remaining coins will always be greater than \(\$1.35\), regardless of which coin he chooses. Therefore, this probability is equal to \(1\), or \(100\%\).

If Dakarai picks only one coin, the only way the remaining coins could have total value less than \(\$2.00\) is if he pulls out the toonie. Thus, the probability is \(\frac{1}{5}=0.2\), or \(20\%\).

There are exactly two coins with total value \(\$0.35\), namely the dime and the quarter. Similarly, there are exactly two coins with total value \(\$3.00\), namely the loonie and the toonie. Since the coins are drawn `at random’, the probabilities of these events must be equal.

Note: The actual probability of each event is \(0.1\) or \(10\%\). This can be illustrated by constructing a tree diagram, or by the following argument. The probability of drawing the dime first is \(\frac{1}{5}\). Then there are only four coins in his pocket, so the probability of drawing the quarter next is \(\frac{1}{4}\). Thus, the probability of drawing the dime and then the quarter is \(\frac{1}{5}\times \frac{1}{4}=\frac{1}{20}=0.05\), or \(5\%\). Similarly, the probability of drawing the quarter and then the dime is \(\frac{1}{5}\times \frac{1}{4}=\frac{1}{20}=0.05\), or \(5\%\). Thus, the total probability of drawing the dime and quarter is \(0.05 + 0.05 = 0.1\), or \(10\%\). A similar analysis can be used to show that the total probability of drawing the loonie and toonie is also \(0.1\), or \(10\%\).