90 cents please
Viktor has to buy some milk. The farmer asks 90 cents for the milk. Viktor has the following coins in his pocket, one of each: 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?
Answer
There are \(\binom{8}{3} = 56\) possible ways to choose 3 coins.
To reach a total of at least 90 cents, Viktor must pick either the 2‑euro coin or the 1‑euro coin (since the sum of any three coins without one of these is less than 90 cents).
If he picks the 2‑euro coin, the other 2 coins can be chosen from the remaining 7 coins in \(\binom{7}{2} = 21\) ways.
If he picks the 1‑euro coin but not the 2‑euro coin (to avoid double counting), the other 2 coins can be chosen from the remaining 6 coins in \(\binom{6}{2} = 15\) ways.
Thus, the total number of favorable outcomes is \(21 + 15 = 36\), and the probability is: