MHB Probability that 12 have purchased yellow gold diamond rings

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The discussion focuses on calculating the probability that 12 out of 20 individuals have purchased yellow gold diamond rings, given that 65% of all diamond rings sold in West Virginia are yellow gold. It is suggested that this scenario can be modeled using a binomial random variable. The relevant formula for this calculation is provided, which includes parameters for the number of trials, successes, and the probability of success. Participants are encouraged to apply the formula with the given values to find the probability. The conversation emphasizes the importance of confirming the method used for this probability calculation.
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Let’s also say that 65% of all diamond rings sold in WV are yellow gold. In a random sample of 20 folks, what is the probability that 12 have purchased yellow gold diamond rings?thank you and I promise this is the last question today.:D
 
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re: probability that 12 have purchased yellow gold diamond rings

Hopefully someone else will confirm this approach, but it seems like you can interpret this to be a binomial random variable. If not then my apologies in advance. Let's say that $P[\text{gold}]=0.65$.

The general formula for a binomial random variable is:

$$P[X=k]=\binom{n}{k}p^{k}(1-p)^{n-k}$$

where $p$ is the success probability, $n$ is the number of trials and $k$ is the number of successes. Can you fill in the information?
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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