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## Probability distributions introduction

Current time:0:00Total duration:3:50

# 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: ## 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.