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?

Question

Ethan Adkins · Accepted Answer

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

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#

Hazel Anglin · Answer

undefined To find the probability that none of the 3 cards will have a winning number, we need to calculate the probability of selecting 3 cards without any of the 2 winning cards.

Total number of ways to select 3 cards from 7 = $ \binom{7}{3} $ = 35

Number of ways to select 3 cards without any of the winning cards = $ \binom{5}{3} $ (selecting 3 cards from the remaining 5 cards that are not winning)

Probability = Number of favorable outcomes / Total number of outcomes
             = $ \binom{5}{3} $ / $ \binom{7}{3} $
             = $ \frac{5!}{3!(5-3)!} $ / $ \frac{7!}{3!(7-3)!} $
             = $ \frac{5 \times 4 \times 3}{3 \times 2 \times 1} $ / $ \frac{7 \times 6 \times 5}{3 \times 2 \times 1} $
             = $ \frac{5 \times 4 \times 3}{7 \times 6 \times 5} $
             = $ \frac{2}{7} $

So, the probability that none of the 3 cards will have a winning number is $ \frac{2}{7} $.