Viktor has to buy some milk. The farmer asks 90 cents for the milk. Viktor knows that he has the following coins in his pocket, one of each coin: 2 euros, 1 euro, 50 cents, 20 cents, 10 cents, 5 cents, 2 cents, 1 cent. What is the probability that – if he grabs three random coins from his pocket – the total value will be 90 cents or higher?
Hint
Use combinatorial analysis. Start with the two euros coin and iterate down to lower value coins.
Answer
Viktor is able to make distinct combinations with 8 coins.
If the two euros coin is involved, it doesn’t matter what the other coins are. So, combinations are sufficient to pay for the milk.
If the one euro coin is involved, another combinations are already sufficient to pay for the milk.
If the two- and one euro coin are both not involved, then we can’t find a combination that’s 90 cents or higher.
So, a total of 36 combinations is sufficient to pay for the milk, out of the 56 possible combinations. Therefore, the probability that Viktor will be able to pay for the milk by grabbing three random coins from his pocket is 36/56.