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
Hi,
Plz explain for second spin how that 2/3 term comes?
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
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
(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
(a) As there are two sections numbered 2, three sections numbered 1, and one section numbered 6, we can find the probability of getting any particular sum by listing all the possible outcomes and their probabilities.
(i) To find the probability that the total of the 2 numbers is 20, we need to consider all the possible combinations. The only possible combination is getting two 10s. Since there is only one way to get this outcome, the probability is 1/6 * 1/6 = 1/36.
(ii) To find the probability that the total of the 2 numbers is 11, we need to consider all the possible combinations. The possible combinations are (5, 6) and (6, 5). Since there are two ways to get this outcome, the probability is 1/6 * 1/6 * 2 = 1/18.
(b) The probability of stopping on a section numbered 2 in one spin is 1/6. Therefore, the probability of not stopping on a section numbered 2 in one spin is 1 - 1/6 = 5/6.
Since we are looking for the probability of stopping on a section numbered 2 on the nth spin, we need to calculate the probability of not stopping on a section numbered 2 in the (n-1) previous spins and then stopping on section 2 in the nth spin.
The probability of not stopping on a section numbered 2 in (n-1) spins is (5/6)^(n-1).
Given that this probability is 16/243, we have the equation:
(5/6)^(n-1) = 16/243.
To solve this equation, we can take the logarithm of both sides:
log[(5/6)^(n-1)] = log(16/243).
Using logarithm properties, we can simplify it to:
(n-1)*log(5/6) = log(16/243).
Finally, we can solve for n by dividing both sides by log(5/6):
n -1 = log(16/243) / log(5/6).
Then, we can add 1 to both sides to get the final value of n.
To answer these questions, we need to understand the probabilities associated with the spinner. Let's go step by step:
(a) Finding the probability that the total of the two numbers is 20:
1. We need to determine all the possible outcomes of spinning the spinner twice.
2. There are 6 sections on the spinner, so the total number of outcomes is 6 * 6 = 36.
3. Now, we need to determine the number of outcomes where the sum of the two numbers is 20.
- The only combination that can give us a sum of 20 is spinning a 2 and an 18.
- As there are two sections with a number 2, we have 2 * 1 = 2 possible outcomes with a sum of 20.
4. Finally, we divide the number of desired outcomes (2) by the total number of outcomes (36).
- 2/36 simplifies to 1/18. Therefore, the probability that the total of the two numbers is 20 is 1/18.
(b) Finding the value of n when the spinner stops on a section numbered 2 with a probability of 16/243:
1. Let's assume n as the number of spins required until the spinner stops on a section numbered 2.
2. The probability of the spinner stopping on a section numbered 2 on any single spin is 1/6 (there are 6 equally likely sections).
3. Since the spinner is spun until it stops on a section numbered 2, the opposite of that event occurring is the spinner not stopping on a 2. So, the probability of not stopping on a 2 in one spin is 5/6.
4. We are told that the probability of stopping on a 2 on the nth spin is 16/243.
5. For n - 1 spins, the probability of not stopping on a 2 is (5/6)^(n - 1).
6. Now, we can set up an equation using the given probability:
- (5/6)^(n - 1) * (1/6) = 16/243
7. Simplifying the equation:
- (5/6)^(n - 1) = (16/243) * (6/1)
- (5/6)^(n - 1) = 16/81
8. Taking the logarithm of both sides, we can solve for n:
- log[(5/6)^(n - 1)] = log(16/81)
- (n - 1) * log(5/6) = log(16/81)
- (n - 1) = log(16/81) / log(5/6)
- n = 1 + (log(16/81) / log(5/6))
9. Using a calculator, we can find the value of n by substituting the values from step 8 into the equation.
- n ≈ 3.249
Therefore, the value of n is approximately 3 (rounded to the nearest whole number).
I hope this clears up the process of finding the probabilities and solving the equations step by step. Let me know if you have any further questions!
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