Probability word problem: 63% of people are happy. Pick 10 people randomly....

MathLearner2016

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63% of people are happy. Pick 10 people randomly.
a) What are the odds that 8 of them are happy?
b) What are the odds that at least 2 of them are happy?
 
For part (a), you might try drawing a tree diagram. When doing so, consider a simpler case where you pick 2 or 3 people. For the first person, you are given the probability that they're happy, but what is the probability they're not happy? As a hint, the probably of something not happening is 1 minus the probability of it happening. Put that information into your diagram. If the first person is happy, what is the probability the second one is? What's the probability they're not happy? Where do you think that information should go on your diagram? Place it accordingly. Now, what if the first person wasn't happy? Does that affect the probability of the second person being happy? Why or why not? Does the first person being unhappy affect the probability of the second person being unhappy? Why or why not? Where do you think you'd put this information on your diagram? Now, consider the third person. Put their information into the diagram. Now, what is the probability of the first and the second people being happy? How did you calculate that? How would you calculate the probability of all three people being happy? Are you seeing a pattern? If not, maybe add a fourth person to the diagram and find the probability of all four being happy.

Now, once you have intuited the pattern (the "formula," if you will) for determining if the first n people are happy, you can use that to help you calculate the probability of exactly 8 of the 10 people are happy. Note that you'll need to consider multiple cases here. Since 8 people are happy, there are two unhappy people. What if the two unhappy people are #1 and #2? What if they're #1 and #3? How many cases, in total, will you need to consider? Is each of the cases equally likely? Why or why not? How does all of this information help you figure out the overall probability of any 8 of the 10 people being happy?

As for part (b), that's actually probably easier than part (a), although it may not look like it at first. As you know, the probability of something happening is 1 minus the probability of it not happening. Since finding the probability of at least 2 people being happy means finding the probability of 2 people, 3 people, 4 people, etc. being happy, it will certainly be simpler to consider the probability of there being less than 2 happy people. So, what is the probability of 0 people being happy? Using your results from part (a) to help you along, what is the probability of 1 person being happy? What is the probability that less than 2 people are happy? Then, what is the probability of at least two people being happy?

If you get stuck, that's okay. But, when you reply, please include all of your work, even if you know it's wrong. Thank you.
 
For part (a)...
png.latex
= 0.02481557802

png.latex
= 0.1369

Answer:
0.1528764

???
 
63% of people are happy. Pick 10 people randomly.
a) What are the odds that 8 of them are happy?
b) What are the odds that at least 2 of them are happy?
a)This is a binomial problem with n=10 and p=.63. If you want r successes (being happy) out of n (# of people randomly selected) then P(exactly r happy people out of n randomly chosen people)= nCr *pr*(1-p)n-r

Now, all you need to do is plug in the values that are given above.

One last note. You said that you want the odds, not the probability. I gave you the probability based on the format of the answer you gave in your last post.

b) p(at least 2 out of 10 are happy)= p(exactly 2 are happy) + p(exactly 3 are happy) + ... + p(exactly 10 are happy).

Might there be a simpler way to calculate the line above?
 
Your calculations above are correct, they just 'aint' the answers to your questions.
You mean because of "odds vs probability"? I meant to say probability. I need to get out of the habit of using them interchangeably.
 
You mean because of "odds vs probability"? I meant to say probability. I need to get out of the habit of using them interchangeably.
Come on, does your results look like you used the formula I stated above or maybe you just used part of the formula?
 
Are you talking about a or b, I'm pretty sure the final answer to a is 0.15.
Both answer.

As I said the work YOU SHOWED is not the answer to a or b.

Show us your work that led to 0.15 and then we will talk about whether it is corrector not.
 
So the probability of nobody being happy is 0.00005; one being happy is 0.0008 and two being happy is 0.0063. Can I use that?
 
So the probability of nobody being happy is 0.00005; one being happy is 0.0008 and two being happy is 0.0063. Can I use that?
What does at least 2 mean to you?

Have you ever said that you got at least an 80% on a test? Did you mean lower than or equal to 80%.

Also it is not polite to just give me those number because I will have to compute the probabilities to see if you're correct. Instead show me the calculations that led to those numbers and that I can tell on the spot if you're correct or not.
 
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What does at least 2 mean to you?

Have you ever said that you got at least an 80% on a test? Did you mean lower than or equal to 80%.
Can't I calculate the probability of 0 or 1 being happy and then subtract that from 1?

P(at least 2) = 1 – P(less than 2)
 
Can't I calculate the probability of 0 or 1 being happy and then subtract that from 1?

P(at least 2) = 1 – P(less than 2)
Yes, but note that you do not need to calculate p(2). Writing p(2) in your last post made me think that you thought at least two meant 0, 1 and 2.
 
Yes, but note that you do not need to calculate p(2). Writing p(2) in your last post made me think that you thought at least two meant 0, 1 and 2.
0 happy: x=0,n=10,p=0.63

(1 - 0.63)^10 = 0.00005

1 happy: x=1,n=10,p=0.63

10 * 0.63 * (1 - 0.63)^9 = 0.0008

Put the two together and subtract from 1:

1 - 0.00085 = 0.99915
 
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