Probability Question

[Alpha Centauri]

Hey guys. Question about probability. I understand that sometimes we need to multiply our probabilities in order to get a final outcome. However, how do we know if we need to add vs just purely multiply these outcomes together to get our final answer? I have attached 2 problem types that exemplify this. I was under the impression that we would just multiply the probabilities of each coin flip since it is 50/50, and then it would be the correct answer. However, we need to add the products of probability for each trial. Why is this?

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[Krentist_72]

I could only open the first one so I'll try to explain.
First of all, as a refresher, multiply when it's AND (for dice, what is the likelihood of rolling two fives in a row: 1/6 AND 1/6 = 1/6 x 1/6) and add when it's OR (what is the likelihood of rolling an even number: 1/6 OR 1/6 OR 1/6 = 1/6+ 1/6 + 1/6).
So by saying that the outcome is three heads and one tail implies that order matters. So you know that the likelihood of 3H and 1T is 1/16, but that's only one possibility of order. You have to take all the ways that you could get 3 H and 1 T and add them because it could be this way OR this way OR this way.
A similar question that I came across in my studies was like: a weatherman predicts 75% rain for Friday, Saturday, and Sunday. What is the probability it rains two of those days? So you would have to realize that it could be Friday AND Saturday, OR, Friday AND Sunday, OR, Saturday AND Sunday, and add up all the possibilities.
If it gives you a specific outcome, what is the likelihood they toss three heads then a tails, or what is the likelihood it rains Friday and Sunday, then the question is giving you a single outcome that you should find-- that is when you don't have to add up all the possibilities.
Sorry that was lengthy but hopefully that helps

 

[Alpha Centauri]

I could only open the first one so I'll try to explain.
First of all, as a refresher, multiply when it's AND (for dice, what is the likelihood of rolling two fives in a row: 1/6 AND 1/6 = 1/6 x 1/6) and add when it's OR (what is the likelihood of rolling an even number: 1/6 OR 1/6 OR 1/6 = 1/6+ 1/6 + 1/6).
So by saying that the outcome is three heads and one tail implies that order matters. So you know that the likelihood of 3H and 1T is 1/16, but that's only one possibility of order. You have to take all the ways that you could get 3 H and 1 T and add them because it could be this way OR this way OR this way.
A similar question that I came across in my studies was like: a weatherman predicts 75% rain for Friday, Saturday, and Sunday. What is the probability it rains two of those days? So you would have to realize that it could be Friday AND Saturday, OR, Friday AND Sunday, OR, Saturday AND Sunday, and add up all the possibilities.
If it gives you a specific outcome, what is the likelihood they toss three heads then a tails, or what is the likelihood it rains Friday and Sunday, then the question is giving you a single outcome that you should find-- that is when you don't have to add up all the possibilities.
Sorry that was lengthy but hopefully that helps

Thanks, so for some probability problems there's no shortcut and you actually have to list out all of the possibilities?

 

[Krentist_72]

Well so if there's 4 flips of the coin, and one head, then it could only be in one of those 4 "positions" so you don't have to list it out, just know to take the answer for one of the trials and multiply it by 4 (aka add it to itself 4 times). Obviously no promises but I would doubt that the DAT would make you recognize a situation where there's like 15 possible outcomes or something lol. I'm pretty sure I got two probability questions and both of them were to find one specific outcome (what is the probability of drawing a red marble then a blue one etc).

 

[Crithu]

This is an example of binomial probability that follows a binomial distribution curve. The equation above is one that is used for these types of problems. It is helpful if you do have a problem where you flip a coin 20 times or some high number like that where is is impossible to write to write out all possible outcomes.

For your problems,
The first one:
4 coins are tossed with the results being 3 heads and 1 tail.

So let's work through the equation. The first part stands for nCr. n is the total number of people and r is the group that we are taking out. In this case, we have 4 coins and we are "choosing" 3 heads. When calculated out, this gives us a value of 4. This number represents all of the possible ways of getting 1 tail and 3 heads. (THHH, HTHH, HHTH, HHHT).

The next part of the equation stands for the probability of success, or in our case, a head. The probability of getting a head is 1/2. r represents the number of times we want this (3). So this part of the equation will read (1/2)^3.

The final part of the equation represents the probability of failure, or in our case, a tail. The probability of this is equal to the probability of not getting a success. (1-1/2). (This is more important to understand in problems with dice for example. If it said the probability of rolling only one 6 on the dice. The probability of a success would be 1/6 and the probability of a failure would be (1-1/6) or 5/6.) n-r represents the number of failures. (4-3)

So our total equation is (4) (1/2)^3 (1- 1/2) ^(4-3). multiply this out and you get 4/16 or 1/4.

Problem 2
A success is a head, and we want only 1 out of 3.

The finished equation would be (3) (1/2)^1 (1- 1/2)^(3-1) and the final answer would be 3/8.

I hope this helps!