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A television commerical claims that 1 out of 5 of "today's marriages" began as an online relationship. Assuming that this is true, calculate the following for eight randomly selected "today's marriages"
A. the probability that at least one began online
B, the probability that 2 or 3 began online
Hi there! Thank you for the opportunity to help you with your question!
First, by this information, we get that the probability of a marriage beginning online is 1/5, and the probability of a marriage not beginning online is 11/5 = 4/5
The probability that at least one began online is the same thing as 1 minus the probability that none of them began online. The probability that none of them began online is given by the product of:
(4/5)*(4/5)*(4/5)*(4/5)*(4/5)*(4/5)*(4/5)*(4/5) = (4/5)^8
So the probability that at least one of them began online is 1(4/5)^8 = 83.2%

The probability that exactly 2 began online is given by:
(8 choose 2) * (1/5)^2 * (4/5)^6 =29.4%
The probability that exactly 3 began online is given by
(8 choose 3) * (1/5)^3 * (4/5)^5 = 14.7%
So the probability that 2 or 3 began is the sum of these two probabilities:
29.4% + 14.7% = 44.1%
Just to clarify, these formulas mean that in order for you to have exactly two people that began online, you need to have the (1/5)*(1/5) and necessarily, all the others must NOT have begun online (4/5)^(82). But, you also need to take the number of possible combinations, because you don't know which two people were married online. Given 8 people ABCDEFGH, the two that began online could be AB, or AC, or AD, and that is why we multiply by (8 choose 2), which is the number of ways you can pick two people out of 8. If you don't know this formula, I recommend the following resource https://en.wikipedia.org/wiki/Binomial_coefficient
Please let me know if you need any clarification. Always glad to help!Hi there! Thank you for the opportunity to help you with your question!
First, by this information, we get that the probability of a marriage beginning online is 1/5, and the probability of a marriage not beginning online is 11/5 = 4/5
The probability that at least one began online is the same thing as 1 minus the probability that none of them began online. The probability that none of them began online is given by the product of:
(4/5)*(4/5)*(4/5)*(4/5)*(4/5)*(4/5)*(4/5)*(4/5) = (4/5)^8
So the probability that at least one of them began online is 1(4/5)^8 = 83.2%

The probability that exactly 2 began online is given by:
(8 choose 2) * (1/5)^2 * (4/5)^6 =29.4%
The probability that exactly 3 began online is given by
(8 choose 3) * (1/5)^3 * (4/5)^5 = 14.7%
So the probability that 2 or 3 began is the sum of these two probabilities:
29.4% + 14.7% = 44.1%
Just to clarify, these formulas mean that in order for you to have exactly two people that began online, you need to have the (1/5)*(1/5) and necessarily, all the others must NOT have begun online (4/5)^(82). But, you also need to take the number of possible combinations, because you don't know which two people were married online. Given 8 people ABCDEFGH, the two that began online could be AB, or AC, or AD, and that is why we multiply by (8 choose 2), which is the number of ways you can pick two people out of 8. If you don't know this formula, I recommend the following resource https://en.wikipedia.org/wiki/Binomial_coefficient
Please let me know if you need any clarification. Always glad to help!Secure Information
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