If the probability of a dancing lady accepting an initation to dance is 0.18, find the expected number of ladies you would have to ask before one accepts.
I think that the answer is: 1(.18)+2(.18)+3(.18)...7(.18)+8(.18)+9(.18)=8.1
*round answers to three signifcant figures*
? The first one could accept. So one.
What certainty of accepatnce are you seeking? 90Percent? 99 Percent? There is a probability you could ask 50 without getting an acceptance. Your method is not right, but the basic question remains, with what certainity?
90% i forgot to say that
To find the expected number of ladies you would have to ask before one accepts, we can use the concept of expected value.
The expected value (E) is calculated by multiplying each possible outcome by its respective probability and then summing them up. In this case, the possible outcomes are the number of ladies you have to ask (1, 2, 3, ...).
Let's break down the calculation step by step:
1. Multiply each outcome by its probability:
- For the first lady, the probability is 0.18, so the contribution to the expected value is 1 * 0.18 = 0.18.
- For the second lady, the probability is also 0.18, so the contribution to the expected value is 2 * 0.18 = 0.36.
- We continue this multiplication for each possible outcome until we reach the desired level of accuracy.
2. Sum up the contributions:
Expected value = 0.18 + 0.36 + 0.54 + ...
Now, let's calculate the expected value of the number of ladies you would have to ask before one accepts:
Expected value = 0.18 + 0.36 + 0.54 + ...
To make the calculation easier, we can notice that each term is a multiple of 0.18. So, we can factor out 0.18:
Expected value = 0.18(1 + 2 + 3 + ...)
This is now a sum of consecutive positive integers. The sum of the first n positive integers is given by the formula n(n+1)/2.
Expected value = 0.18(n(n+1)/2)
Now, we need to find the value of n. We want to find the expected number of ladies you would have to ask before one accepts, so we stop when a lady accepts. In this case, n would be 1 less than the number of ladies because we stop counting after the accepting lady.
Therefore, n = 1 less than the expected number of ladies.
We can solve for n by plugging in the expected value of ladies (E) into the formula:
E = 0.18(n(n+1)/2)
8.1 = 0.18(n(n+1)/2)
Now, we can solve for n:
8.1 = 0.09(n^2 + n)
89.99 = n^2 + n
n^2 + n - 89.99 = 0
Now, we can solve this quadratic equation using the quadratic formula or by factoring.
To get an approximate value, let's use the quadratic formula:
n = (-1 ± √(1^2 - 4(1)(-89.99))) / (2(1))
n = (-1 ± √(1 + 359.96)) / 2
n = (-1 ± √(360.96)) / 2
n = (-1 ± 18.999) / 2
Since we cannot have a negative number of ladies, we can discard the negative solution.
n ≈ (-1 + 18.999) / 2 ≈ 9.999 / 2 ≈ 4.999 ≈ 5 (rounded to the nearest whole number)
Therefore, the expected number of ladies you would have to ask before one accepts is approximately 5.