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# Probability with discrete random variable example

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

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• 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? •   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.
• 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) •   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``
• 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? • 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... :) • I don't understand how the instructor calculated the probability of buying 2 packs at . I don't understand the logic. Can someone explain? • 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.
• 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 • 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.    