help with stats hw

Question

A standard deck of cards contains 52 cards. One card is selected from the deck.

​(a) Compute the probability of randomly selecting a five or seven. ​

(b) Compute the probability of randomly selecting a five or seven or nine.

​(c) Compute the probability of randomly selecting a ten or club.

Answer 1

I love playing cards! In a standard deck of cards, there are fours "suits": Clubs, Diamonds, Hearts and Spades. There are 13 cards in each suit: Ace, 2, 3, ...10, Jack, Queen and King. A total of 4X13=52 cards

The probability of selecting a 5 or a 7:

We have 4 fives, (one in each suit) : five of clubs (5C), five of diamonds (5D), five of hearts (5H), five of spades(5S).

Similarly there are 4 sevens: 7C; 7D; 7H and 7S

So 8 of the 52 cards are 5's or 7's.

Probability of a 5 or 7 is 8/52 or 2/13

Make sense?

Part b is similar: 5, 7 or 9

4 fives, 4 sevens, 4 nines 4+4+4=12

Probability of a five, seven or nine = 12/52 or 3/13

Part c is a little tricky. Probability of a ten or a club.

Well there are 13 clubs, and 4 tens but one of the tens is the 10 of clubs, and we don't want to "double count" that card.

13 (clubs)+ 4(tens) - 1(ten of clubs)=16

P(ten or Club)=16/52

So FUN!

Answer 2

1 out of 13 ways to get a 5 4/52 = 1/13

1 out of 13 ways to get a 7

2 out of 13 ways to get 5 or 7 8/52=2/13

3 out of 13 ways to get 5, 7 or 9 12/52 = 3/13

1 out of 4 ways to get a club

1 out of 13 ways to get a 10

there is an overlap though with the 10 of clubs in both sets

total ways to get a club or 10 is 4 out 52 plus 1 out of 4, but subtract 1/52 for the 10 of clubs to avoid over counting. 4/52 = 1/13 + 1/4 - 1/52 = 4/52 + 13/52 - 1/52 = 16/52 = 4/13

13 ways to get a club, 4 ways to get 10. 13+4 = 17 subtract the 10 of clubs 17-1=16

that gives 16 out of 52 cards that will give you a ten or a club 16/52=4/13

Unless when you say "or" you mean only a ten or only a club, but not both a 10 and a club? "or" can be used in either way, inclusive or exclusive, to mean A or B or both. But also sometimes it means A or B but not both.