Probability with discrete random variable example | Random variables | AP Statistics | Khan Academy

Example analyzing discrete probability distribution.

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[artashestadevosyan]

How is the 2nd one being calculated ? Why 0.8x0.2 ? Why do not to the same with 3 ?

[jorgemercent]

Sal, i think you made a mistake :/

P(X=4)=0.8*0.8*0.8*0.2=0.1024
P(X>/= 2)= P(X=2) + P(X+3)+ P(X+4) = 0.16 + 0.128 + 0.1024 = 0.3904 (Should be the correct answer)

Total Probability = P(X>/= 1) + P(0) = 0.2 + 0.3904 + (0.8x0.8x0.8x0.8) = 1
and not
Total Probability = P(X>/= 1) = 0.2 + 0.3904

You forgot P(X=0)?

[TasyaAdzkiya]

I think the probability of Hugo getting the card that he wants is 0.2 in the first purchase because they said that each pack has probability 0.2 of containing the card Hugo is hoping for.

If he didn't get it in the first pack purchase, he can buy another pack (because he has the money to buy up to 4 packs). The probability of him getting the card that he wants in the second pack purchase is 0.16 (because if he didn't get it in the first one, we know that the probability of him getting the card is 0.2, of course it also means that the probability of him not getting it is 0.8), so 0.8 (probability of not getting it on the first purchase) x 0.2 (new hope or the probability of getting the card on the third pack that he is about to purchase) = 0.16. We use the general multiplication rule of probability because we're dealing with a dependent probability. Also, I quoted this from article on Khan Academy titled "The general multiplication rule": "When we calculate probabilities involving one event AND another event occurring, we multiply their probabilities. In some cases, the first event happening impacts the probability of the second event."

We assume that his only goal is to get that one card (the card of his favorite player). Suppose that he doesn't have the reason to purchase the cards packs other than getting the card. So, if he purchase another pack, it definitely means that he didn't get it in the previous pack. The probability of getting it in the third purchase is 0.8 (he didn't make it in the first purchase) x (0.8 he didn't make it in the second purchase) x 0.2 (new hope or the probability of getting the card on the fourth pack, it would be the last hope because he can only afford up to 4 packs).

The probability of him getting it on the fourth card = 0.8 (he didn't make it on the first purchase) x (he didn't make it on the second purchase) x 0.8 (he didn't make it on the third purchase). We don't multiply it with 0.2 (0.2 is the new hope or the probability of getting the card for each pack, 0.8 is the probability of NOT getting the card for each pack) because he can't afford to buy more packs. He can only afford up to 4 packs, we only multiply 0.2 if there we plan to purchase another pack (because the 0.2 is the probability of getting the card in the upcoming purchase), which we will not do because Hugo can't afford more than 4 packs. Even if he didn't get the card that he wants on the last purchase (in this case, the fourth purchase), he can't purchase another pack anymore because he can't afford it. So, he stops there.

If you add up all of the probabilities, they will add up to 1 (or 100%, since 100 percent is just 100/100). That is because it is established in the first place that he can only buy up to 4 packs, that's why it makes sense that it will add up to 100%. This is what we call discrete probability. A discrete probability distribution counts occurrences that have countable or finite outcomes (this definition is from Google, sorry). In this case, Hugo can only purchase up to 4 packs, so it's finite. That's why if we add up all of the probabilities, it adds up to 1.

If he don't have any constraints (for example, he doesn't have budget limit and he can buy it continuously), you can see that every purchase, if we add up all of the probabilities, it's more closer and closer to 100% or 1 but never 1. So, for all probabilities to be able to reach 1 or 100%, it has to be fixed/established how many packs you can purchase to make the probabilities add up to 100%.

For example, if we don't stop at 4 packs, the probability of getting the card in the fourth purchase is 0.8 (he didn't make it on the first purchase) x 0.8 (he didn't make it on the second purchase) x 0.8 (he didn't make it on the third purchase) x 0.2 (new hope or the probability of getting the card on the next purchase), and so on. You keep doing that continuously because in that case where you don't have limits, you will do it infinitely and the probabilities will never add up to 100% or 1.

3:23 "This seems like a high probability, there is more than 50% chance that he buys 4 packs. But, you have to remember that 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." means that there is 20% probability of him stopping there after the first purchase because he got the card on the first purchase (because we assume that he only buys the pack of cards just to get that one card and there's no other reason for him to purchase any packs if he got it on the previous purchase), 16% probability of him stopping there after the second purchase because he got the card on the second purchase, 12.8% probability of him stopping there after the third purchase because he got the card on the third purchase, and there is 51.2% probability of him stopping there after the fourth purchase because he got the card on the fourth purchase AND there is also the probability that he only stop there (even if he didn't get the card at all) because he just can't afford any pack of cards anymore. So, the difference of the fourth purchase compared to the first, second, and third purchase is that the on the fourth purchase, there is also probability that he only stops there because he can't afford any pack of cards anymore, while if he stops at the first, second, or third purchase, that is definitely because he got the card and doesn't have to purchase any pack of cards anymore (assuming that he will not stop at less than 4 purchase if he hasn't got it). So, that's why the probability of him stopping there at the fourth purchase is relatively high. Please correct me if I'm wrong and sorry for the bad grammar!

[mohammedhamed]

why is the probability of the 2nd is .2*.8 ??

[erolxtreme]

51% of Hugo to stop there, EVEN IF he will not find the card he was searching for.

[RobertCramer-lh]

Wouldn't it be great if the money you had to spend would control when you are guaranteed to have success? I would go spend it all on lotto tickets right now.

[yash]

3:04 so, Sal gave up the old trusted on-screen Ti84 calculator and got a mac right!! 🤣🤣

[anangtruong]

Hi guys, for some of you who have the question about whether the P(4) should be 0.128*0.8 or ) or 0.512, I think ) .512 is the correct answer but it will be incorrect if the question was to define X as the numbers of card Hugo buys until he win knowing that he only can buy four times.

[WAKEUPLIONS]

What is 0.8 plz explain me how 0.8 comes😢

[sakshamgarg]

I think there is an error in this video:

The P(X=4) should not be 0.512. Instead, it shall be 0.1024. Explanation:
P(X>=4) = P(X>3) = 1 - P(X<=3) = 1 - (0.2 + 0.16 + 0.128) = 0.512
P(X = 4) = P(X>=4) * P(finding a card in a pack) = 0.512 * 0.2 = 0.1024