Statistics - Probability of Random Events

[Quincy]

Homework Statement

1. A bus arrives at a station every day at a random time between 1:00 PM and 1:30 PM.

a) What is the probability that the person has to wait exactly 15 minutes for the bus?

b) What is the probability that the person has to wait between 15 and 20 minutes for the bus?

2. Suppose you will be given a random subset of five cards from a standard deck of 52 cards.

a) What is the probability that you’ll get exactly one ace?

b) What is the probability that you’ll get at least one ace?

Homework Equations

The Attempt at a Solution

1 a) 1/30 -- I figured, since the bus arrives at any random time during the 30 minutes, it has 1/30 chance of arriving at any given minute.

b) 1/6?... I'm not really sure how to approach this one...

2 a) (4/52)*(48/52)*(48/52)*(48/52)*(48/52) = 0.05585?

b) (4/52) + (4/52) + (4/52) + (4/52) + (4/52) = 20/52?

 

[dacruick]

1a) Assuming that the person gets there exactly at 1pm that is true. If they get there past 1, and the bus hasn't come yet things change. I think you are right though.

b) you could do two things, you could find the probability of the person waiting more than 20, and the probability of the person waiting less than 15, and then subtract those two from 1.
or you could do it directly. Once again, assuming the person arrives at 1pm exactly, I think you are right

2a) almost right! Remember that the second card isn't mutually exclusive from the first. As goes the rest of them. What this means is that a card previously drawn from the deck affects the probability of the next card being drawn.

2b)This one is quite wrong. But I think I've given you the tools in my above suggestions to do this properly. Do you have any ideas?

 

[Quincy]

2a) almost right! Remember that the second card isn't mutually exclusive from the first. As goes the rest of them. What this means is that a card previously drawn from the deck affects the probability of the next card being drawn.

2b)This one is quite wrong. But I think I've given you the tools in my above suggestions to do this properly. Do you have any ideas?

2a) (4/52)*(47/51)*(46/50)*(45/49)*(44/48) = 0.0549? -- but wouldn't this be the probability that an ace is picked exactly once AND is first to be picked?

2b) (4/52) + (3/51) + (2/50) + (1/49) = 0.19515?

 

[dacruick]

Those are both still wrong, but this question is fairly difficult. I'm not even sure I remember how to do it properly so I'm going to try and work this out right here.

Lets try and do this out with 2 cards being drawn, and see if we can extend the logic out to 3 then 5.

2a) with 2 cards: (4/52)*(48/51) (I do 48/51 because no aces can be picked, and there are only 3 left, something you missed)
So once again, that is the way an ace is picked up first only. So the other way that could happen is if we do (48/52)*(4/51). Let's add those together and see what we get.
(4/52)*(48/51) + (48/52)*(4/51) = .1448 This seems like a plausible answer so let's try and check out answer to confirm that its right. We can do this by finding the amount of ways we don't get any aces, and the amount of ways that we get 2 aces.
No Aces: (48/52) * (47/51) = .8506
Two Aces: (4/52) * (3/51) = .0045
So P(1Ace) = .1448
P(0Ace) + P(2Ace) = .8551 = 1 - P(1Ace)

Hopefully you followed that with no problems. I'd like you to try and do the 3 drawn cards exercise for me, then we can see if we can come up with a nice equation and extrapolate to 5 cards without having to do it out manually.

EDIT: I'm also ignoring 2b for now, after 2a is finished that won't be a problem for you.