Compound Probability of Independent Events Coins 52 Playing Cards

in this video we're going to focus on compound probability but particularly of independent events so let's say we wish to find probability of two events occurring let's say event and event if these events are independent then the probability of these two events occurring is simply the product of the probability that event will occur and then event will occur so let's keep that in mind now let's focus on this question what is the probability of getting heads twice after flipping coin so we need to get heads on the first try and then heads on the second try so the probability of getting these two events is going to be the probability of getting heads on the first try times the probability of heads on the second try so what is the probability of getting heads just by flipping with coin well if we write out the sample space there's only two options either we get heads or tails so the probability of getting heads is going to be this one option out of the two possible options so it's one out of two now for the second event the probability will be the same it's also one out of two 1 Time 1 is 1 2 * 2 is four so the probability of getting two heads is one out of four now to confirm this answer let's write out the sample space for flipping two coins we can get heads and then heads or we can get heads then Tails or we can get Tails first and then heads on the second try or we can get two tails so notice that we have one event that is our desired event out of potential of four events so that's why the probability is one out of four out of the four events only one leads to successful outcome the outcome that we want and so that's one way in which you can confirm that answer now what about Part what is the prob probility of getting heads Tails heads after flipping coin three times so what you need to know is that the order matters in this problem but getting the answer is pretty straightforward so HTH this is going to be the probability of getting heads on the first try times the probability of getting tails on the second try times the probability of getting heads on the third try and so for each of these events the probability of them occurring is 1/2 so this is basically 1 12 raised to the 3 power 2 the 3 is 8 and 1 the 3r is 1 so the probability of this happening is 1 out of eight now let's confirm it let's write out the sample space for flipping three coins so we can get all three heads or it could be heads heads and then Tails or it can be or HTT HH and so forth so these are the eight possibilities that we can get when flipping three coins so notice that there's only one out of these eight possibilities that leads to the event that we want and so that's why the probability is one out of eight now let's move on to part what is the probability of getting two tails and one heads so does the order matter in this problem so looking at the eight options that we have here which one contains two tails and one heads I'm going to highlight it in Red so this has two tails and one heads so there's this one and that one as well so therefore the probability of getting either HTT or th HT or tth is three out of eight because we have three favorable outcomes out of eight possibilities that will lead to this event so you need to be careful where if the order matters or if it doesn't so for this problem the order matters but for the second problem the order doesn't matter so it's good to list out all the potential events that would lead to this particular situation number two in standard 52 card deck what is the probability of selecting queen well before we go into this question let's talk about some things you need to know so in standard 52 card deck you don't have any Jokers the first type of card that you have is the red diamond the second card is the red hearts the third one which is going to be black but can't use black on black screen so I'm going to use white instead the third one represents the Spades which looks something like this my drawing's not perfect but what you can do is go to Google Images type in standard 52 card deck and you'll see all the cards in the 52 card deck but this is supposed to be black instead of white just keep that in mind the last type is the clubs which looks something like this let me see if can draw that better but it's something like that you'll see it when you search it on Google Images now for each of these four different types of cards you have the ace and then you have the numbers two all the way to 10 after 10 you have the Jack and then the queen and then the king so this right here represents 13 cards so what this means is that there's diamonds 13 Hearts 13 Spades and 13 clubs 1 * 4 gives us the total of 52 cards so there's two types of cards that are 13 * 2 is 26 so there's 26 red cards and there's 26 black cards so these are some things to know if you're dealing with probability question using cards so let's begin you may want to write those things down as notes to answer these questions coming up so in standard 52 card deck what is the probability of selecting queen so how many queens are there so we have the Queen of Diamonds the Queen of Hearts the Queen of Spades and the Queen of Clubs so there's only four types of Queens because there's four different types of cards so there's four Queens out of total of 52 cards 4 is 4 * 1 52 is 4 * 13 so we could cancel four and so the answer is going to be 1 out of 13 1 / 13 is basically 07 692 that's red answer so if you multiply that by 100 the probability is approximately 7.69% now what about Part what is the probability of selecting black card that is greater than three but less than or equal to 9 so let's write out the list of numbers that fit these characteristics so let's say that is basically the number of card that we're going to pick it's greater than three but less than or equal to 9 so this does do not include three so if we Write the sample space this is going to be 4 5 6 7 8 9 now you might be thinking okay so the probability is going to be 6 out of 52 but keep in mind there two types of black cards you have the clubs and you have the Spades so there's six black clubs with these numbers and there's six Spades with those same numbers so really there's 12 cards that is between that's greater than three less than or equal to 9 that is black card because you got the clubs and the Spades so it's 12 out of 52 12 is 4 * 3 52 is 4 * 13 so if we cancel four the probability is going to be three out of 13 this is the answer for Part now let's move on to our last question or rather the last part in this question part what is the probability of selecting red queen on not red queen but red King on the first try and then diamond card on the second try without replacement now this problem is little bit tricky so go ahead and try it now what we don't want to do is this we don't want to say it's going to be the probability of selecting red King times the probability of selecting diamond because the first event is going to affect the second event for instance let's say if we want to find probability of selecting red King there's two red Kings out of 52 cards you could have the red diamond or the red king of harps now the probability of selecting diamond without replacement we took out card so there's 51 cards left now how many diamonds are left now the number of diamonds that are left depends on which Red King that you selected if we selected red King of Hearts then there's 13 diamonds left including the Red King of Diamonds but if on the first try we selected the Red King of Diamonds then there's 12 diamonds left so these are dependent events the second event depends on the first event so we can't use this equation for this particular situation now another formula that we could use is when you have an or situation the probability of selecting or is equal to the probability ility of plus the probability of we're going to have to use this so event is going to represent the first situation that is let's say if we select red King of Hearts and then on the second event will be selecting diamond event will be selecting red King of Diamonds and then the second event will be selected under the diamond card so let's start with the first event this is going to be the probability of getting the Red King of Hearts times the probability of getting diamond card so we know that this four red Kings mean there's four King cards two of them are red the diamonds and the hearts so to get red King of Hearts there's only one red king of hearts out of 52 cards so the probability is one out of 52 now to get diamond card after that there's once we take out the Red King there's going to be not 52 cards in deck 51 cards because this is without replacement we're not putting the Red King of Hearts back into the deck so now we have 51 cards to choose from 13 of those cards are diamonds so now let's do the math 52 is basically 4 * 13 so we can cancel 13 and 4 * 50 is 200 so 4 * 51 is going to be 204 you just add another four so the probability of getting red King of Hearts and then diamond card is one out of 204 now let's focus on the other events event so this is if we get red King of Diamonds let me put the probability of getting red King of Diamonds times the prob of getting diamond card so there's only one red King of Diamonds out of the 52 cards now this is without replacement once we take that card out there's 51 cards left now there's 13 Diamond cards we took out one of the diamond cards so that's there's 12 Diamond cards left over so the probability for the second event is going to be the 12 diamonds out of the 51 cards that remains 52 we're going to say it's 4 * 12 is 4 * 3 51 is 17 * 3 so we're going to cancel four and we're going to cancel three so what we have on the bottom is 13 * 17 13 * 17 is 221 so the probability of selecting red King of Diamonds on the first try and then domond card on the second try without replacement is one out of 221 now let's put all of this together so the probability of getting red King followed by diamond card is the sum of the probability of getting red King of Hearts because that is red King followed by diamond card plus the probability of getting red King of Diamonds which is still red King followed by diamond card we had to break up this problem into these two situations because the probability for those two situations are different so this is going to be 1 over 204 and then the other one is 1 over 221 so we need to get common denominators in order to add these two fractions let's multiply the second fraction by 204 and the first one by 221 204 * 221 is 4584 and then we could add 221 + 204 which is 425 now we can probably simplify our answer somehow these are some large numbers so this is Let's take step back perhaps it's better if we simplified it from the beginning now 221 that's basically we said that was 13 * 17 204 is 51 * 4 Now 51 we can break that into 17 * 3 so 17 * 3 * 4 that'll give us the original 204 so now we need to get common denominators we already have 17 this has 12 that's 3 * 4 so we need to multiply the fraction on the right by 12 12 the fraction on the right has 13 which the left one or the one on the left doesn't have so we're going to multiply the fraction on the left by 13 over 13 so this should give us combinator mean wow common denominators that are less in value than what we had before so 13 * 17 * 3 * 4 that's 2,652 13 * 17 * 12 is the same thing 2652 so now we can add 13 + 12 which is 25 so this right here is the final answer 25 over 2,652 that is the probability of getting red King on the first try and then diamond card on the second try without replacement