ALGEBRA
posted by Mike on .
A single card is selected from an ordinary deck of cards. The sample space is shown in the following figure. Find the following probabilities. (Enter the answers either as fractions or as decimals rounded to three places.)
P(five of clubs) = Incorrect: Your answer is incorrect.
P(five) = Correct: Your answer is correct.
P(club) = Incorrect: Your answer is incorrect.
P(jack) = Correct: Your answer is correct.
P(spade) = answer is incorrect.
P(jack of spades) =
P(five and a jack) =
P(five or a jack) =
P(heart and a jack) =
P(heart or a jack)

5 of C = 1/52
5 = 4/52 =1/13
club = 1/4 = 13/52
jack = 4/52 =1/13
spade same as club
J of S, same as 5 of C
You only took one card. It can not be 5 and jack
five or jack 1/13 + 1/13 = 2/13
heart and jack is jack of hearts 1/52
Heart OR jack
draw Venn diagram
There are 13 hearts
There are 4 jacks, three of which are not hearts. The jack of hearts is the intersection of the sets, careful not to count it twice.
So out of the 52 cards
13 are Hearts and 3 more are jacks that are not hearts
total = 13+3 = 16
so
16/52 = 4/13
heart or jack = 1/4 + 3/52 
A single card is selected from an ordinary deck of cards. The sample space is shown in the following figure. Find the following probabilities. (Enter the answers either as fractions or as decimals rounded to three places.) A single card is selected from an ordinary deck of cards. The sample space is shown in the following figure. Find the following probabilities. (Enter the answers either as fractions or as decimals rounded to three places.)

p(5 or club)