maths-probability

The diagram shows a spinner with six numbered sections.(where it has number 2 twice, # 1 triple & # 6 once )
Some of the sections are shaded.
Each time the spinner is spun it stops on one of the six sections.
It is equally likely that it stops on any one of the sections.
(a)The spinner is now spun twice. find the probability that the total of the 2 numbers is
(i)20 (ii) 11

(b)The spinner is now spun until it stops on a section numbered 2.
The probability that this happens on the nth spin is
16/243.
Find the value of n.
kindly show the steps

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  1. Since all you can get is 1, 2 or 6
    there is no way to get a total of 20 or 11 with only 2 spins.
    looks like either a typo or a flaw in the question.

    2nd part.
    Spin until you get a 2 ....

    getting a 2 on
    1 spin = (1/3)
    2 spins = (2/3)(1/3) = 2^1 / 3^2
    3 spins = (2/3)^2 (1/3) = 2^2 / 3^3
    4 spins = (2/3)^3 (1/3) = 2^3 / 3^4
    ...
    n spins = 2^(n-1) / 3^n

    so 2^(n-1) / 3^n = 16/243
    2^(n-1) / 3^n = 2^4 / 3^5

    n = 5

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  2. ard so i have a question. so im always on this web site looking for answers yes i do "cheat" if thats what you wanna call it. but half the time im confused, wth is the point in this website? is it to help you or is it for answers cuz i stay getting 100s on my tests from answers on here. i do do my work i just check the answers on here and if ion know a answer best believe this is my first stop but. like what is the point in the website.??? and if you got the answers for Lesson 3 Sample Spaces Unit 6 quick check post the answers cuz i need help. Thankkkss

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  3. Hi,
    Plz explain for second spin how that 2/3 term comes?

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  4. Hey , the question says that the spinner should stop in section number 2 on the nth spin. This means in the rest of the spins before, the spinner should stop on any number apart from 2.
    so the probability that it will stop once on 2 (for the last spin/ spin n)= 1/3
    and the probability it will stop on anything else apart from 2 is = 2/3

    so probability = 1/3 (last spin )* 2/3^n-1 ( as this is the number of times the spinner lands on anything part from 2)
    so 1/3 * 2/3^(n-1) = 16/243
    n = 5

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  5. (2/3)^(n-1) * 1/3 = 16/243
    (2/3)^(n-1) = 16/81
    2^(n-1)/3^(n-1) = 2^4/3^4
    n-1 = 4
    n = 5

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