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Probability with discrete random variable example

AP.STATS:
VAR‑5 (EU)
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VAR‑5.A (LO)
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VAR‑5.A.1 (EK)
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VAR‑5.A.2 (EK)
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VAR‑5.A.3 (EK)
CCSS.Math:
Example analyzing discrete probability distribution.

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  • aqualine ultimate style avatar for user Giorgio
    But... if you calculate P(2) as 0.8 (or the 'missing chance' in the first pack) times 0.2 (the 'finding chance' in the second pack), obtaining 0.16 probabilities of him buying 2 packs, I will suppose I have to do 0.8 * 0.8 * 0.2 obtaining 0.128 to calculate the probabilities of him buying three packs.

    So...why isn't it 0.8*0.8*0.8*0.2 = 0.1024 probabilities of him buying four packs?

    Because here you have all the potential probabilities of him buying more than four packs if he could...I guess, but I'm not grasping it clearly. How can I calculate that? Or what is a clearer way of thinking about it?
    (47 votes)
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    • blobby green style avatar for user Harrison Sona Ndalama
      So the key to understanding this solution is to realise that there are actually 5 possible outcomes here. i.e. X=1 because the first card is his favorite
      X=2 because the 2nd card is his favorite
      X=3 because 3rd card is his favorite
      X=4 because 4th card is his favorite
      and
      X=4 when 4th card is NOT his favorite but he has to stop anyway (because of no money)

      Thus using the reasoning you supposed, the probabilities are calculated as follows:
      P(X=1) = 0.2
      P(X=2) = 0.8*0.2= 0.16
      P(X=3) = 0.8^2*0.2 =0.128
      P(X=4) = 0.8^3*0.2 +0.8^4 =0.512

      The key is to realize that X=4 is composed of two possible outcomes i.e. he either gets his favorite card on the 4th try.... or he doesn't but still has to stop after the 4th try.
      (73 votes)
  • starky sapling style avatar for user Fadel Hilmy
    I think you have a mistake in the way you explain there.

    P(X=4) = 0.8*0.8*0.8*0.2 = 0.1024

    But don't forget that there is still P(X=5), P(X=6), P(X=7), ... . It's just that our main character here (Hugo) can only buy 4, but it's still a probability that he finally get his fav player card on the tenth try right? And all of that P(X=x) , with x is a positive real number, will sum up to 1. Lets ask Sal to further explain this in another video. SO we know that :

    P(X=1) + P(X=2) + P(X=3) + ... = 1

    The question is P(X>=2), so you will add P(X=2) + P(X=3) + P(X=3) + ... . By using the equation of the sum of all possibilities, we can get :

    P(X=1) + P(X=2) + P(X=3) + ... = 1
    P(X=2) + P(X=3) + ... = 1 - P(X=1)
    (6 votes)
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    • leafers ultimate style avatar for user Phil P
      There's no mistake in the video.

      You're misinterpreting P(X=4) to mean the probability that Hugo gets the card he wants in the fourth pack (and not the first 3). That isn't correct. P(X=4) is the probability that Hugo buys 4 packs, regardless of whether the 4th pack contains the card or not.

      P(X=5), P(X=6), etc will all be zero, because Hugo can't buy more than 4 packs.

      In other words, P(X=4) is the probability that the Hugo gets the card in the 4th pack plus the probability that he doesn't get the card at all.

      Here's a tree that might help:
        -----------------
      | |
      | P(card) | P(not card)
      | = 0.2 | = 0.8
      | |
      P(X=1) -----------------
      = 0.2 | |
      | P(card) | P(not card)
      | = 0.2 | = 0.8
      | |
      P(X=2) = -------------------
      0.8 * 0.2 | |
      = 0.16 | P(card) | P(not card)
      | = 0.2 | = 0.8
      | |
      P(X=3) = -------------------
      0.8^2 * 0.2 | |
      = 0.128 | P(card) | P(not card)
      | = 0.2 | = 0.8
      | |
      P(card in pack 4) P(didn't get card)
      = 0.8^3 * 0.2 = 0.8^4
      = 0.1024 = 0.4096
      \ /
      \ /
      ---------------
      |
      P(X=4) =
      0.1024 + 0.4096
      = 0.512
      (111 votes)
  • blobby green style avatar for user Muhurthana84
    I have been watching this for the past 1 hour. I do not understand the part 0.8 . 0.2 where X=2. Any idea how that became 0.16?
    (6 votes)
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  • blobby blue style avatar for user Nalini Singh
    I don't understand how the instructor calculated the probability of buying 2 packs at . I don't understand the logic. Can someone explain?
    (3 votes)
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    • leaf green style avatar for user George D.
      Allow me to explain. Hugo has a 0.2 probability of getting the card he wants on any given card. Conversely, he had a 0.8 (80%) probability of not getting his desired card. So, the chances of Hugo stopping to shop after two cards is the two numbers multiplied. 0.2*0.8=0.16, which is the probability that he leaves with his card after buying a second card. Greetings from 2021, take care.
      (5 votes)
  • male robot hal style avatar for user Nikos Menoudakis
    Hi everyone! I have been puzzled in the same way on this exercise. My opinion is that the confusion is due to a sort of mistake in the "declaration" of the random variable: based on the description of the problem and the table given, I believe that X is not "the number of packs that Hugo buys" but "the number of packs that Hugo stops buying at"!! I mean, if you picture it this way there is no need to further explain P(1) or even P(4) for that matter. Anyway, it worked nicely for me... :)
    (5 votes)
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  • mr pink red style avatar for user sulaiman  alomayri
    There:
    Hugo wants to find his favorite is basketball card and he is willing to pay a pack of cards and check of it has his favorite player card. He will do this only 4 times., so P(X) is the probability that he might find his card in the X pack he bought. FIRST YOU NEED TO FIND THE PROBABILITY SUCCEEDING OF EACH PACK.
    = Probability of missing * probability of succeeding

    P(1)= Probability of succeeding= 0.2
    Missing= .8

    To find P(2), we have to assume he missed on P(1)
    So: .8 (missed probability from p(1) * .2 ( Succeeding P(2) )
    =.16

    To find P(3), we have to assume he missed on P(1) AND P(2)
    and So on
    =.8 * .8 * .2( succeed on P(3)
    = .128

    P(4) same process:

    =.8* .8*.8.2
    =.1024
    (5 votes)
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  • winston baby style avatar for user Yash Singh
    This is a geometric model. The probability is 0.8 x 0.8 x 0.8 x 0.2 = 0.1024. The 0.8s represent a failure. The 0.2s represent getting his favorite card. The probability should decrease gradually. There should be another section saying that he never reached his goal. That section should say 0.4096 which is 1- all the other sections.
    (4 votes)
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  • blobby green style avatar for user everett10238
    Hopefully, this comment helps anyone who is confused with the problem. It is best that you read the whole comment because you may become confused if you do not. For anyone who did not know, when finding the probability of two or more events you multiply the first probability to the next probability; you follow this step consecutively with all given probabilities, multiplying each probability to the next probability and including all previous probabilities in an expression. The key to this problem is recognizing that 20% or 0.2 applies to both the probability of Hugo purchasing one pack—P(1) = 0.2—and the chance of him obtaining his favorite card in any pack; the given probabilities are the probabilities of Hugo purchasing the corresponding number of packs—P(2) = 0.16, P(3) = 0.128, & P(4) = ?—not the probability of Hugo obtaining his favorite card if he purchases packs collectively. In any given pack there is a 80% or 0.8 chance of it not having Hugo’s favorite card and a 20% or 0.2 chance of it having the card; TWENTY PERCENT AND EIGHTY PERCENT, REGARDING THE CHANCE OF HUGO’S FAVORITE CARDS, ARE FACTORS IN THE EXPRESSIONS FOR CALCULATING PROBABILITY; I assume that the reason you use the card’s 80% and 20% probability to calculate the probability for the number of packs after one is because they are the only constant feature of this problem. (If anyone can better explain my assumption please do so.) If Hugo does not obtain the card in pack one, he will purchase another pack, so you would calculate the probability of him purchasing a second pack by 0.8 x 0.2; you would say 0.8 x 0.2 since Hugo “fell into” the 80% with pack one (which acts as a factor when calculating the probabilities of each pack ) and he is hoping to “land in” in the 20% (another factor when calculating probability) with the next pack; this is why the probability for pack two is 0.16. Remember that every time you calculate the probability for a succeeding pack, you include 0.8 as a factor for each previous pack since calculating the probability for a succeeding pack means that Hugo did not obtain his favorite card in the previous pack. So, the expression for pack three would look like 0.8 x 0.8 x 0.2 = 0.128. Actually, none of the information above was needed to solve the problem; it just helps you understand the given information more. The problem—Find the indicated probability [of] P( X >_ 2 )= _ [I tried my best to show the eaquation in text] )— indicates that you are to find the collective probability of two packs, three packs, and four packs, excluding the first pack in the process. So, since we are expected to know that all the possibilities aggregately equal 100% or 1 and since we are only finding the probability of two packs and greater, you exclude the probability of pack one and subtract it from 1, which leaves you with 80% or 0.8. So, automatically, the probability (answer) for P ( X >_ 2 ) = 0.8. If you subtract the probabilities of pack two and pack three each from 0.8, you find that you are left with 0.512, the probability of Hugo ultimately buying four packs. Some people confuse 0.1024 as being the probability of four packs; however, since Hugo’s budget limits him to 4 packs at most, 0.1024 would not be the answer; THIS WAS JUST EXTRA WORK, SO DO NOT MENTALLY CONFUSE THIS WITH YOUR ANSWER; As Sal said, this seems like a high percentage, but the limited amount of packs Hugo can purchase makes him more likely to buy a fourth pack if he did not obtain his favorite card in the first three attempts. A logical way to look at this is that Hugo would possibly be more induce to buy a fourth pack since he did not obtain the card in his three previous purchases and since his budget allows him to purchase one more pack. This comment mostly addresses the confusion from what I saw in the Questions that most people meet when watching the video. The major problem most people have is interpreting the problem, so maybe I or someone else will post a comment deconstructing the question and what you have to do.
    (2 votes)
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  • blobby green style avatar for user abrolaryan17
    what if we have to find the probability distribution. How would we find that.
    (2 votes)
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  • duskpin seedling style avatar for user Monique Swart
    This question is a bit confusing to me because of the way it's phrased. Are we looking for the probability that he has to buy another card, of the probability that he finds his favourite card? Wouldn't the probability of him finding the card he's looking for go up, the more cards he buys? But if it's the probability of him having to buy another pack, it would be 0.8 at X = 1 so it's not that either.

    I'm not sure if I misunderstood the question.
    (2 votes)
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Video transcript

- [Lecturer] Hugo plans to buy packs of baseball cards until he gets the card of his favorite player, but he only has enough money to buy at most four packs. Suppose that each pack has probability 0.2 of containing the card Hugo is hoping for. Let the random variable X be the number of packs of cards Hugo buys. Here is the probability distribution for X. So it looks like there is a 0.2 probability that he buys one pack, and that makes sense because that first pack, there is a 0.2 probability that it contains his favorite player's card, and if it does, at that point he'll just stop, he won't buy any more packs. Now what about the probability that he buys two packs? Well, over here they give it a 0.16, and that makes sense, there is a .8 probability that he does not get the card he wants on the first one, and then there's another .2 that he gets it on the second one. So 0.8 times 0.2 does indeed equal 0.16. But they're not asking us to calculate that, they give it to us. Then, the probability that he gets three packs is 0.128, and then they've left blank the probability that he gets four packs. But this is the entire discrete probability distribution, because Hugo has to stop at four, even if he doesn't get the card he wants at four on the fourth pack, he's just going to stop over there. So we could actually figure out this question mark by just realizing that these four probabilities have to add up to one. But let's just first answer the question, find the indicated probability, what is the probability that X is greater than or equal to two? What is the probability, remember, X is the number of packs of cards Hugo buys. I encourage you to pause the video and try to figure it out. So let's look at the scenarios we're talking about. Probability that our discrete random variable X is greater than or equal to two, well, that's these three scenarios right over here. And so what is their combined probability? Well you might want to say, hey we did figure out what the probability of getting exactly four packs are, but we have to remember that these all add up to 100%. And so this right over here is 0.2 and so this is 0.2, the other three combined have to add up to 0.8. 0.8 plus 0.2 is one, or 100%. So just like that we know that this is 0.8. If for kicks we wanted to figure out this question mark right over here we could just say that look, have to add up to one. So, we could say, the probability of exactly four is going to be equal to one minus 0.2 minus 0.16 minus 0.128. I get one minus .2 minus .16 minus .128 is equal to 0.512. Is equal to 0.512, 0.512. You might immediately say, wait, wait, this seems like a very high probability, there's more than a 50% chance that he buys four packs, and you have to remember, he has to stop at four. Even if on the fourth he doesn't get the card he wants he still has to stop there. So, there's a high probability that that's where we end up. There's a little less than 50% chance that he gets the card he's looking for before that point.