## anonymous 5 years ago A student guesses the answers to 6 questions on a true-false quiz. Find the probability that the indicated number of guesses are correct: no more than 2(hint:no more than 2 means exactly 0 means exactly 1 or exactly 2)

1. anonymous

put X = number he gets right. the probability he gets any one right is the probability he gets one any one wrong is $\frac{1}{2}$ to get six wrong in a row multiply $(\frac{1}{2})^6$ to get one right and 5 wrong it is $\dbinom{6}{1}(\frac{1}{2})^6$ $\dbinom{6}{1}$ is the number of ways to choose on item from a set of 6, so it is 6

2. anonymous

Thank you!!

3. anonymous

$(\frac{1}{2})^6=\frac{1}{64}$ $6\times \frac{1}{64}=\frac{6}{64}$ $\frac{1}{64}+\frac{6}{64}=\frac{7}{64}$ $1-\frac{7}{64}=\frac{57}{64}$

4. anonymous

oops ignore the last line. just $\frac{1}{64}+\frac{6}{64}=\frac{7}{64}$

5. anonymous

man i am off. you also have to compute the probability that he gets two right. forgot about that one.

6. anonymous

it is $\dbinom{6}{2}\frac{1}{64}$ and $\dbinom{6}{2}=15$

7. anonymous

so the correct answer is $\frac{1}{64}+\frac{6}{64}+\frac{15}{64}=\frac{22}{64}$

8. anonymous

sorry my mistake. not reading carefully.

9. anonymous

its okay! thank you! i got the answer correct :)

10. anonymous

great!

11. anonymous

are you good with pascal's triangle and other probabilities?

12. anonymous

sure got a question?

13. anonymous

lets see...what is the probability of getting 3 red lights at a traffic stop, when looking at five lights. assuming the common changes are red and green

14. anonymous

assuming these probabilities are each ${1}{2}\ for each light and assuming they are independent? 15. anonymous yeahh 16. anonymous a rather broad assumption. so this is like saying if you flip a coin 5 times what is the probability you get 3 heads and one tail, yes? 17. anonymous oh yeah, you could actually put it that way! 18. anonymous then it is just \[\dbinom{5}{3}(\frac{1}{2})^5$

19. anonymous

$(\frac{1}{2})^5=\frac{1}{32}$ do you know how to compute $\dbinom{5}{3}$?

20. anonymous

no? haha i am not very great with math

21. anonymous

do you know how to write the first few levels of pascal's triangle?

22. anonymous

yeah i know how to do that

23. anonymous

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

24. anonymous

there it is down to level 5 to compute each entry start at 1, then add the top two adjacent numbers.

25. anonymous

so to get the next level i would start at 1, then say 1+5=6 5+10=15 10+5=15 5+1=6 and so the next level would be 1 6 15 15 6 1

26. anonymous

okay...got it!

27. anonymous

but let us just say without writing the triangle i wanted to compute $\dbinom{7}{3}$ the number of ways to choose 3 from 7. i would make a fraction and put in the top $7\times 6\times 5$ in the bottom $3\times 2$tp get $\frac{7\times 6 \times 5}{3\times 2}$ this has to be a whole number so cancel before multiplying.

28. anonymous

$\frac{7\times 6 \times 5}{3\times 2}=7\times 5=35$

29. anonymous

$\dbinom{10}{4}=\frac{10\times 9 \times 8 \times 7}{4\times 3\times 2}$

30. anonymous

again a whole number so cancel first and multiply last.

31. anonymous

$\dbinom{10}{4}=10\times 3\times 7=210$

32. anonymous

oh wow, okay i get it a little better now

33. anonymous
34. anonymous

more than you want to know

35. anonymous

okay i have another question that im having a hard time with

36. anonymous

shoot

37. anonymous

Suppose that you ask three friends to go to the mall. each one has a 0.80 chance of saying yes. What is the probability that all three friends will say they can go to the mall with you?

38. anonymous

ok that one is easy. $(.80)^3$

39. anonymous

assuming independence multiply the probabilities

40. anonymous

oh so its 0.512

41. anonymous

cool!

42. anonymous

you got it.

43. anonymous

wow thank you! are you good with the binomial theorem? haha

44. anonymous

or with the Fibonacci Sequence?

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