Probablities!!! help me lol im desperate!?

Question

I have answers for the probs but im not sure if they are correct though. So please help?

1.) find the probability for the experiment of tossing a coin 3 times & the sample space for it is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}<<< THATS NOT THE PROB THATS THE INSTUCTION :}
- find the probability of getting at least two heads.
i got 4/3 but im not sure if thats correct.

2.) A college student is preparing a course schedule for the next semester. The student may select one of two mathematics courses, one of three science courses, and one of five courses from the social sciences and humanities. How many schedules are possible?
& the answer i got was 120.

3.) You can answer any 10 questions from a total of 12 questions on an exam. In how many different ways can you select the questions?
& i got 66.

Thanks.

Answer 1

1.) Only 4 of the 8 possibilities in the sample space have at least 2 heads. So the probability is 4/8 = 1/2. Note that if you get over 1 for your probability you are incorrect, 4/3 can't be the answer.

2.) The answer is 2 * 3 * 5 = 30 schedules. The choices are independent, so you just multiply the number of possibilities for each choice together to get your total number of possibilities.

3.) Ok, your answer here is just 12 choose 10 = 12!/(10!2! = 12*11/2 = 66. You are correct.

Answer 2

All probabilities are between 0 (impossible) and 1 (certain). You can't have a probability of 4/3.

So for (1) there are 8 possibilities, three of which have exactly two heads, and 1 of which has three heads. So the probability is 3/8 + 1/8 = 4/8 = 1/2.
.
For (2), how did you get 120? Hmmm. There are 2 * 3 * 5 possibilities, which is 30.

For (3), yes, 66. The easiest way to see this is that you just have to select any 2 not to answer. So 12 * 11 / 2 = 66.
.

Answer 3

For starters, a probability must be between 0 and 1, so 4/3 is not correct.

Probability is defined as (# of desired outcomes) / (# total outcomes).

Here, there are 8 total outcomes (you enumerated them above). How many of these have at least two heads: HHT, HTH, THH, and HHH. That's 4. So, the probability is 4/8 = 1/2.

In general, if you want to flip a coin N times and get M heads, the formula is (N choose M) × (prob of Heads)^M × (prob of Tails)^(N-M).

In this problem, we have 2 heads out of 3. (3 choose 2) × (1/2)^2 × (1/2)^(3-2) = 3 × (1/2)^3 = 3/8. Next, we need 3 heads out of 3. (3 choose 3) × (1/2)^3 × (1/2)^(3-0) = 1/8. Add these together to get 4/8 = 1/2.

Answer 4

One thing you should know about probability is that it can never be above one.

1) Your answer is 1/3 more than 1 so it's definitely wrong. You flip it 3 times and each time the probability of getting a head is 1/2. You have a 1/8 chance of getting a heads. Since you want to get two, I assume the chance is halved again. The chance of getting 2 heads is 1/16.

2) He has to choose one of each so he has 2 possibilities for math, 3 for science, and 5 for anthropology or whatever.
2*3*5=30 possible schedules. If you can choose 5 from the other one, since you said AND, there would be 150 choices.

3) This would be a combination problem, I believe. You're picking 10 out of 12 and the order of picking them doesn't really matter.
12!/(10!2!)
12*11/2! = 6*11 = 66
Yep, you're correct.

Answer 5

1. There are 4 ways to get 2 Heads. The sample space is 8.
So probability is 4/8 = 1/2.

2. 2*3*5=30

3. 12C10 = 12!/2!*10! 12*11/2=66
You got this one right, and I believe it was the hardest one.

Answer 6

1) You have 8 possible outcomes.
Out of this 4 of them are favorable.
That's 4/8 = 1/2
Probability can never exceed 1.

2) The answer is simply 2*3*5 = 30
not 120

3) Your answer of 66 is correct.
Answer = 12C10 = 12*11 / 2 = 66

Answer 7

1) 2/3 2)it is 30 2science corses times 2 math courses times 5 social sciences gets you 30 possible schedules 3) is kind of confusing i think it's 120 but im not sure. any way good luck I havent done much math lately so might want a second opinion