Yves and some friends are playing a fair game in which 18 cards are dealt to 6 players. One of the cards is a queen. The player who receives the queen goes first. What is the probability that Yves will go first?
a) 1/18 b) 1/3 c) 1/6 d) 1/13
each of the six players have an equal chance.
1/6?
To find the probability that Yves will go first, we need to determine the number of favorable outcomes (cards received by Yves) and divide it by the total number of possible outcomes (all possible ways the cards can be dealt).
Let's calculate it step by step:
Step 1: Determine the total number of possible outcomes.
Since there are 18 cards and 6 players, we can use the concept of permutations to determine the number of possible outcomes. The number of ways to deal 18 cards to 6 players is 18P6, which can be calculated as:
nPr = n! / (n - r)!
18P6 = 18! / (18 - 6)!
= 18! / 12!
Step 2: Determine the number of favorable outcomes.
Yves will go first if he receives the queen. There is only one queen card, so Yves needs to receive it. The remaining 17 cards can be dealt to the other players in any order. Hence, the number of favorable outcomes is 17P5, which can be calculated as:
nPr = n! / (n - r)!
17P5 = 17! / (17 - 5)!
= 17! / 12!
Step 3: Calculate the probability.
Finally, we can find the probability that Yves will go first by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Number of favorable outcomes / Total number of possible outcomes
= 17P5 / 18P6
= (17! / 12!) / (18! / 12!)
Simplifying this expression yields:
Probability = (17! * 12!) / (18! * 12!)
= 1/18
Therefore, the probability that Yves will go first is 1/18.
Therefore, the correct answer is (a) 1/18.