2 you possess a standard deck of playing cards n 52 first a...

(2) You possess a 'standard deck of playing cards' (n = 52). First, (a) identify the probability of selecting a club, diamond, or heart. Second, (b) calculate the probability of selecting a club, spade, heart, or face card. Identify (c) the probability of selecting (in sequence) an ace and then a red king (assuming that the first card was replaced), and (d) the probability of selecting (in sequence) a club and then a red seven (assuming that the first card was not replaced). Still using one 'standard deck of playing cards' (n = 52), consider Event A [a diamond or an ace] and Event B [a black card]. Calculate (e) the probability of Event A. Next, (f) identify the probability of Event B. Then, (g) calculate the probability of Event A or Event B, i.e., P(A or B). Now, (h) identify the probability of Event A, given that Event B occurred, i.e., P(A|B). Finally, (i) using two separate 'standard decks of cards,' compute the probability of Event A occurring in the first deck and of Event B occurring in the second deck.

Transcript

00:01 This problem involves many parts centered around the probability of cards.

00:06 So let's talk about cards in general before we start.

00:09 There are 52 cards in a standard deck of cards.

00:12 Some of them are black cards.

00:16 There are 26 of them.

00:18 And some of them are red cards.

00:21 There are 26 of those.

00:23 The black cards are found in two suits.

00:27 We've got clubs and we've got spades.

00:30 There are 13 of each, and the red cards we have hearts and diamonds.

00:43 There are 13 of each of those.

00:46 Keep in mind that each of the suits, the 13 cards, consist of an ace, a king, a queen, a jack, a 10, a 9, an 8, a 7, a 6, a 5, a 4, a 3, and a two, and the ace king queen of each suit would be classified as face cards.

01:15 So we are now ready to begin our probabilities now that we have this laid out.

01:22 So for part a, we want to determine the probability of selecting a club or a diamond or a heart.

01:34 So whenever you're doing probabilities separated by or, you can use the addition rule so we could say it's the probability of a club plus the probability of a diamond plus the probability of a heart.

01:48 There are 13 out of 52 that are clubs.

01:53 13 out of 52 are diamonds and 13 out of 52 are hearts.

01:59 Now, 13 out of 52 can be expressed as an equivalent fraction of one -fourth, so it's technically one -fourth plus one -fourth plus one -fourth or three -fourths.

02:13 For part b, we want to determine the probability of selecting a club or a spade or a heart or a face card.

02:30 So again, we'll set it up the probability of a club plus the probability of a spade, plus the probability of a heart, plus the probability of a face card.

02:44 But this scenario is not mutually exclusive.

02:55 There is some overlap.

02:57 There are some face cards that are some clubs and face cards that are spades and face cards that are hearts.

03:03 So therefore, some of our cards have been counted twice.

03:07 So we will have to subtract out the probability of a club being a face card and the probability of a spade being a face card and the probability of our heart being a face card.

03:29 So the probability of a club would be 13 out of 52.

03:35 Of a spade is 13 out of 52.

03:38 A heart is 13 out of 52.

03:42 And there are 12 face cards out of the 52.

03:46 But there are three face cards that are clubs.

03:50 So we'll have to subtract them out because we counted them twice.

03:54 There are three face cards that are spades.

03:58 And there are three face cards that are hearts.

04:03 And if we were to add these four, we would get 51 out of 52, and we would be subtracting 9 out of 52, yielding 42 out of 52, and an equivalent reduced fraction would be 21 out of 26.

04:28 For part c, we want to calculate the probability of selecting in sequence an ace, then a red king.

04:49 So when we're doing probability of one followed by another, this becomes conditional probability.

04:55 So we're going to determine the probability of an ace, and we're going to multiply it by the probability of a red king.

05:05 Now, there is replacement happening in between picks.

05:17 So the probability of an ace, there are four cards out of 52.

05:22 We get one, we put it back, and there happened to be two red kings out of the 52.

05:30 So i could simplify or write equivalent fractions, four out of 52 is the same thing as 113th.

05:38 Two out of 52 is the same as 126th.

05:43 And if i multiply those, i get an overall probability of 1 over 338.

05:53 Part d, we want to determine the probability of selecting in sequence a club and then a red 7.

06:14 And this time, it's without repeat.

06:17 Placement.

06:23 So therefore, we're going to start off with the probability of getting that club, and we'll multiply it by the probability of getting that red seven, given that we know a club is already out of the deck.

06:38 So the probability of a club is going to be 13 out of 52, and red sevens, there are two of them, but we know a club has already been pulled from the deck and has not been put back.

06:56 So there's 51 cards that are still in the deck.

07:01 So we could multiply these.

07:07 And we would get one -fourth as a more simplified fraction times 2 out of 51, which would be 2 out of 204 or 1 out of 102 as a probability.

07:29 Now from here, we're going to now define event a and event b and continue with some probabilities.

07:38 So event a is going to be selecting a diamond or an ace.

07:51 And event b is going to be selecting a black card.

08:00 So in part e, we want to determine the probability of event a.

08:08 So that's the same thing as saying the probability of a diamond or an ace.

08:18 So we'll do the probability of a diamond plus the probability of ace...