PROBABILITY MATH HELP!?

I need an explanation/walkthrough on how to do these two questions!

1) If you roll a pair of dice 41 times, what is the probability that both die will show a 1 exactly 3 times?

2) A multiple choice quiz has 7 questions on it, and each question has 3 options. If you did not study and had to guess each answer, what is the probability you passed the test? (Passed = got, 4, 5, 6, or 7 questions right).

Update:

And this next one is just something I m confused about.

***A supplier estimates that 0% of their parts are defective. If you place an order for 30 products, what is the probability that at least one is defective?

→ Wont it just be zero then????

5 Answers

Relevance
  • 5 years ago
    Favorite Answer

    Well,

    edit : wait a minute, mademoiselle : I am checking with the calc.

    1) n=41 independent experiences, p=1/36 is the proba of success at each experience represented by RV Xi

    (i from 1 to n=41)

    P(Xi = 1) = 1/36 success and P(Xi = 0) = 1 - p = 35/36

    N : number of successes

    N = X1 + X2 + ... + X41 has a BINOMIAL distr. B(n,p)

    P(N = k) = C(n,k) p^k (1-p)^(n-k)

    therefore, here :

    P(N=3) = C(41,3) (1/36)^3 * (35/36)^(41-3)

    = 41*40*39/(3*2*1) * 35^38/36^41

    and I hope you can calculate by your own ... !! ;-)

    my calc gives :

    P(N=3) = 0,07833220621275444346985755828065

    P(N=3) # 7.83%

    2) same scheme : 7 indep. experiences, p = 1/3 of success at each question

    n=7 , p=1/3

    so I suppose you "pass" if you have a majority of correct answer, ok?

    therefore :

    P( pass the test) = P(N >=4)

    calculate :

    P(N=0) = C(7,0) (1/3)^0 (2/3)^7 = 1 * 1 * (2/3)^7 # 0.05853

    P(N=1) = C(7,1) (1/3)^1 (2/3)^6 = 7 * 2^6/3^7 # 0.2048

    P(N=2) = C(7,2) (1/3)^2 (2/3)^5 = (7*6)/2 * 2^5/3^7 # 0.3073

    P(N=3) = C(7,3) (1/3)^3 (2/3)^4 = (7*6*5)/(3*2*1) * 2^4/3^7 # 0.2561

    calculate :

    P(N <= 3) = P(N=0) + P(N=1) + P(N=2) + P(N=3) ==> gives you P( N <= 3)

    my calc gives :

    P(N <= 3) = 0.8267

    and, of course :

    P( pass the test) = P(N >=4) = 1 - P(N<=3) gives you the answer.

    P( pass the test) = 0.1733 # 17.3% ... not so bad ... !! ;-)

    3) A supplier estimates that 0% of their parts are defective

    therefore :

    whether you place 30 or 30,000,000 parts : NONE is defective !!

    the thing is that : this is pure theory and does not exist in real life ... ;-)

    et voilà, mademoiselle !! ;-)

    hope it' ll help !!

    PS: if you want good answer, do not forget to give Best Answers..... to me, or anybody else, or course provided the answer deserves one !

    this is to encourage people to answer !! ;-)

    • Commenter avatarLogin to reply the answers
  • Null
    Lv 4
    5 years ago

    In the first question, I'll use binomial distribution.

    Let X be the number of rolls with getting (1,1). for the binomial distribution parameters are "n" & "p". We have n=41, now we need the probability of getting (1,1) and it is 1/36.

    X~BIN(n=41,p=1/36)

    the pmf of the binomial distribution is as follows

    P[X=x]=C(n,x)*(p^x)*(q^(n-x)) .... Here C denotes combination & q=1-p=1-(1/36)=35/36

    we are asked P[X=3] = ? ... So replace x as 3 in the pmf

    P[X=3]=C(41,3)*((1/36)^3)*((35/36)^38)

    the result is approximately 0,08 ...

    P[X=3] ≈ 0.08 .... http://www.wolframalpha.com/input/?i=C%2841%2C3%29...

    2nd one can be solved by again binomial distribution. I will leave it to you. (Hint: You should consider the each part seperate and add them up, like P[X=4]+P[X=5]+P[X=6]+P[X=7]. Think about what the random variable should be. Naming the random variable is a very important thing in statistics and probability related things.

    • Commenter avatarLogin to reply the answers
  • cidyah
    Lv 7
    5 years ago

    n = 41 (number of times the pair of dice is thrown)

    p = 1/36 (probability that both roll a '1' in a single throw

    x= number of throws showing (1,1) out of 41 throws

    P(x=3) = 41C3 (1/36)^3 (35/36)^(41-3)

    P(x=3) = 41C3 (1/36)^3 (35/36)^38 ----> binomial probability

    = (10660) (1/36)^3 (35/36)^38

    = 0.0483

    • Commenter avatarLogin to reply the answers
  • 5 years ago

    1) probability of getting two 1s on a pair of dice is 1/36, so probability of not is 35/36

    (1/36)^3 * (35/36)^40 * 43!/(40!3!) = .0857

    2) probability of getting a question right = 1/3 (1/3)^4(2/3)^3(7!/3!4!) + (1/3)^5(2/3)^2(7!/5!2!) + (1/3)^6(2/3)(7!/6!) + (1/3)^7

    • Commenter avatarLogin to reply the answers
  • How do you think about the answers? You can sign in to vote the answer.
  • 5 years ago

    ugh thank you all sosososo much

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.