Question

Three cards are selected at random from a group of 7. Two of the cards have been marked with winning numbers. What is the probability that none of the 3 cards will have a winning number?

Answer 1

`P("not pick a winner")=10/35`

Explanation:

We are picking 3 cards from a pool of 7. We can use the combination formula to see the number of different ways we can do that:

`C_(n,k)=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

`C_(7,3)=(7!)/((3!)(7-3)!)=(7!)/(3!4!)=(7xx6xx5xx4!)/(3xx2xx4!)=35`

Of those 35 ways, we want to pick the three cards that do not have any of the two winning cards. We can therefore take the 2 winning cards from the pool and see how many ways we can pick from them:

`C_(5,3)=(5!)/((3!)(5-3)!)=(5!)/(3!2!)=(5!)/(3!2!)=(5xx4xx3!)/(3!xx2)=10`

And so the probability of not picking a winning card is:

`P("not pick a winner")=10/35`