P(face card U Jack)

Question

Solving Probability

Answer 1

First, let's translate this math language to English. This is asking for the probability of pulling a face card or a Jack. That U-looking symbol means "or." The biggest tricky part for this problem is that these are not really independent events, because there are some face cards that are Jacks.

 

When dealing with probability problems involving "or," and when the events are non-independent like this one, remember this equation called the "addition rule."

 

P(A or B) = P(A) + P(B) - P(A and B)

 

Let's call P(A) the probability of pulling a face card. The face cards are Jacks, Queens, and Kings. Because there are four suits, there are a total of 12 face cards. The total number of cards in a deck is 52, so the probability of pulling a face card is 12/52. That fraction can be reduced to 3/13. We'll call this P(A).

 

P(B) will be the probability of pulling a Jack. There are only four jacks in a deck, so the probability for a Jack is 4/52. That fraction can be reduced to 1/13.

 

P(A and B) is asking how many cards are both face cards and Jacks. Well, the only cards that satisfy both those conditions are the Jacks. Each Jack card is a face card and also a Jack. The probability for this one, therefore, is the same as the probability for P(Jack), which was 1/13.

 

Plug all these terms in the addition rule equation.

 

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = 3/13 + 1/13 - 1/13

P(A or B) = 3/13

 

Those two terms at the end subtract themselves because look at it this way. We're asked what the probability is of pulling a face card or a Jack. That's pretty much asking "What's the probability of pulling ANY face card?" This is because Jacks themselves are face cards.

 

I hope this helps, and good luck with your math class!