Question

Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?

Answer 1

Answer:

P(at least 1 pair)  = $\frac{17}{33}$

Step-by-step explanation:

Bill has total of 12 cards.

The total probability is 1.

P(at least 1 pair) + P(no pairs) = 1

So, P(at least 1 pair) = 1 - P(no pairs).

Now let's find the probability that no pairs occurs.

1st card has taken. Now we are going to take the second one and see the probability that will not match with the first one.

Now the total number of cards = 11

The number of favorable outcomes (does not match) = 10

P(2nd card does not match the first) = $\frac{10}{11}$

Now we have only 10 cards.

Favorable outcomes = 8 (Does not match)

P(3rd card does not match the first 2 cards) =  $\frac{8}{10}$

=  $\frac{4}{5}$

P(4th card does not match the first 3 cards) =  $\frac{6}{9}$

=  $\frac{2}{3[tex]

So, P(no pairs) = \frac{10}{11}.\frac{4}{5} .\frac{2}{3}$}[/tex]

P(no pairs) = $\frac{16}{33}$

Now to find the probability of at least 1 pair.

P(at least 1 pair) = $1 - \frac{16}{33}$

P(at least 1 pair)  = $\frac{17}{33}$