A college has 1000 students. Twenty-five of these students are Asian-Americans. If you draw a random sample of 10 students from the student body, what is the probability that at least one of the students in the sample is an Asian-American?

What about if you draw a random sample of 50 students?

A blackjack player has been losing all night. On the very last hand of the evening, he decides to bet everything he has left thinking he is "due".

Is this a good idea?

Why or why not?

I'm no proability whiz, but if 2.5% of the students are Asian, then I'd think whatever the number of your draw, then 2.5% of that draw would be Asian.

As for Blackjack, no way. The cards of that last hand are dependent on what cards have been played since the last shuffle.

Related: If you flip a coin 9 times and get heads all 9 of them then your odds are still 50-50 that the next flip will be heads.

I was a probability whiz in high school :)

This is one of those questions where you should look at it the other way around, because it's easier to figure out that way. If you took 10 of the 1000 students, what are the odds that those 10 are _not_ asian-american?

Well, start with one student:

975/1000 - no problem.

The next student?

974/999 - okay

the odds that both of the first two are not asian-american is the product of those two probabilities:

975/1000 * 974/999

What then, are the odds that at least one of the students is asian-american? Well, it's one minus the odds that neither one is asian-american:

1 - (975/1000 * 974/999)

You should be able to extend this to 10 students, or 50 students, or as many as you want.

An interesting note on this: When your probability teacher makes a bet that at least two people in your class have the same birthday, don't take it. In a class of 25 people, the probability that two people share a birthday is:

1 - 365! /( 340! * 365^25) = 0.56

Wow!

Your second question has to do with something called "independent trials". If each hand of blackjack is truly random, then preceding hands have no bearing on the current hand. Like rickj's comment about flipping the coin - if it comes up heads a million times in a row, the probability of getting a tails the next flip is still 50%.

Of course, in real life, previous trials do tell you things about the current test. If a coin flips heads 10 times in a row, maybe it's because the coin is weighted so it preferentially lands heads side up. If the blackjack player has lost all night long, maybe it's because the dealer is cheating, or because he's just a really crappy blackjack player. Maybe he accidentally sat down at the poker table, or worse, pai-gow!

The blackjack player only "feels" they are due to win. Only a professional card counter has any valid information what is "due" in the next hand.

While "counting" the cards that have been dealt they will know exactly if they are the favorite in the next hand and my how much. Very powerful information but very difficult to master. That is why I and other counters (once discovered) are not allowed to play high level blackjack in any casino world-wide.

Two Free odds/strategy cards are available on my website.

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The first problem is classic binomial probability.

For 'at least one' find the probability of none and subtract from 1.

1-C(10,0)(\frac{1}{40})^{0}(\frac{39}{40})^{10}={0.2 24}

You could also sum them up from 1 to 10:

\sum_{k=1}^{10}C(10,k)(\frac{1}{40})^{k}(\frac{39} {40})^{10-k}\approx{0.224}

Probability that none of the students is Asian-American
=

Number of ways of selecting 10 out of 975 divided by number of ways of selecting
10 out of 1000

= 975C10 / 1000C10 = (975! 990!)/(965! 1000!) = 0.7754

Probability that at least one is Asian American = 1 - 0.7754 = 0.2246

Proceed similarly in case of 50:

Probability = 1 - 975C50/1000C50 = 1 - 0.272970524

= 0.727029476