Question 1

At 0:15, Sal begins to list all the possible outcomes for flipping 3 coins. Is there any way to figure out the number of outcomes without listing them or using a chart?

Accepted Answer

2 outcomes for one event; 3 events. Therefore 2^3 = 8 outcomes.

Question 2

How many people are needed in a room so that the probability of two people sharing the same birthday is roughly one-half?  I know the answer is 23, but I do not know how.

Accepted Answer

a: Possible birthday permutations: 365^n
b: Permutations with no shared birthdays: 365! / (365-n)

Probability of n people NOT sharing birthday:
b / a = 365! / (365-n)! / 365^n = 365! / (365-n)! / 365^n

n=23:
365! / (365-23)! / 365^23 = 0.493

Probability of one or more shared birthday between 23 people: 
1 - b/a = 0.507

I _think_ that's an acceptable way to do it, but I've been wrong before.

Question 3

How am i supposed to solve this with a tree diagram?

Accepted Answer

draw t, h in the first row, each of those branch out into t and h, so there will be 2 ts and 2hs in the second row, then those branch out into 2 ts and hs each, so there are 4ts and 4hs in the third row

Question 4

Thank you for the awesome lecture :) I suddenly got a question on my mind. I guess it's about probability and possiblity? Okay. Let's say we have to find a genious student. We had total 100 students, but couldn't find the genious one. Than if there are more than 100 students, like 500 or something? Wouldn't the possibility to find the genious student can be higher? But I think this theory contradicts with the probabilitical method. Whoa I'm getting more confused as I write it.

Accepted Answer

You are on the right track. Given that there are genius students with in the population, the larger the samples size of the population you take, the greater the possibility (that is to say, it is more probable it is) that your sample will have a genius student with in it.
So, with a smaller sample size, eg 100 students, it is less likely you will find a genius student within it than if your sample size was 500 students, but that does not mean the finding a genius student in the smaller size is impossible, it could happen, but is just less likely to happen.
Keep it up!
Keep thinking and asking questions!

Question 5

Why is THH and HHT considered two different possible outcomes? Are they not the same outcome?

Accepted Answer

When 3 coins (or any other specified number of coins) are tossed, every sequence of heads and/or tails is equally likely. Each possible sequence occurs with probability (1/2)(1/2)(1/2) = 1/8 (or more generally with probability 1/(2^n) if n coins are tossed). However, not every combination of heads and/or tails without regard to order is equally likely. So in this type of situation, order of heads and/or tails is important for computing probabilities.

Have a blessed, wonderful day!

Have a blessed, wonderful day!

Question 6

What if we change this to 5 coins?

Accepted Answer

See Probability is nothing but   *Expected Events  / All Possible Events*
So here if we flip 5 coins then all possible events =  *2 * 2 * 2 * 2 * 2* =   *2^5* = *32*
Expected event is getting exactly two head, if you know how combination works then you may find that event of getting exact 2 heads in 5 flips is nothing but  *5C2* =  *10*
So Probability of getting exactly two heads in 5 flips =  *10/32* OR  *5/16*

Question 7

So I get that a deck of cards may be 52!. But, as I play some other card games I was wondering how I'd work out the possible hand combinations if say my deck contains duplicate cards?:
 ( do I just minus them from n? ) i.e (n- *minus duplicates*)!/_k_!*1/_k_  ?

Accepted Answer

If you have a deck of n cards, there are n! ways of arranging them.

Now if you want to figure out how many different hands you could get with 5 cards in hand, using a deck of 52 cards for instance, you'd use the "choose function" Which will come soon after in the probability section, but here is a "sneak peak". the choose function takes 2 parameters, n = how many things to choose from (here 52, all the cards) and how many do we want to pick (here 5 for our example)
choose(52,5) = 52! / (52-5)! * 5! = 2,598,960

To generalize it, let's say N is total of cards to choose from, and K the number of cards you'll have in your hand
choose(n,k)= n! / (n-k)! * k!

Question 8

if from a 100 sided polygon 3 points are chosen at random. Then what is the probability that the chosen points form a right angled triangle? I know the answer is 1/33 but I can't solve it.

Accepted Answer

I might be wrong about this, but this is how I would prove it:
Let's start with the first point. That point could form 100 different angles with the next point and the horizontal axis: approximately 0, 3.6, 7.2, 10.8.......up to 180. The reason that they are all multiples of 3.6 is because 3.6 is approximately equal to 360(a full circle of degrees) divided by 99(the number of points on the polygon).
So the first point could be anywhere. The second point has to form a complementary angle with the first one, the x-axis, and the third point to make a right angle. Example: if point A forms an angle of about 10.8 with point B and the x-axis, then point B  _has_ to make the angle 349.2 with point C and the x-axis. The probability of point B making that exact angle is 1/99, since it has to connect with point C, which could be randomly at any of the 99 points. So the probability would be 1/99, except that we have to remember that this layout could start with any point, not just point A. It could start with point B or point C. That triples it's chances of happening, so the final probability would be `1/33.`
Hopefully this was clear enough. A diagram would be clearer, but I can't write one here. :)

Get ready for Precalculus

Course: Get ready for Precalculus > Unit 8

Probability with counting outcomes

Want to join the conversation?

Video transcript