Suppose that you randomly draw one card from a standard deck of 52 cards. after writing down which card was drawn, you place the card back in the deck, shuffle the deck, and draw another card. you repeat this process until you have drawn 12 cards in all. what is the probability of drawing at least 8 diamonds? for the experiment above, let x denote the number of diamonds that are drawn. for this random variable, find its expected value and standard deviation.

Probability of getting 8 or more diamonds = 0.00278151 Mean and standard deviation = 3 and 1.5 Since you're replacing the card and shuffling, each event is independent and has a probability of 1/4 of success and 3/4 of failure. You succeed if you pick 8, 9, 10, 11, or 12 diamonds. For picking 8 diamonds, there are 12!(8!4!) ways of picking exactly 8 diamonds and the probability of each of those possibilities is 0.25^8 * 0.75^4, so the total probability of exactly 8 diamonds is 12!(8!4!)*0.25^8*0.75^4 = 0.002389848 Continuing with the same logic, the probabilites for 9, 10, 11, and 12 diamonds is 12!(9!3!)*0.25^9*0.75^3 = 0.000354052 12!(10!2!)*0.25^10*0.75^2 = 0.0000354052 12!(11!1!)*0.25^11*0.75^1 = 0.00000214577 12!(12!0!)*0.25^12*0.75^0 = 0.0000000596046 And the total probability is the sum of those 5 terms. Giving 0.002389848 + 0.000354052 + 0.0000354052 + 0.00000214577 + 0.0000000596046 = 0.00278151 This is a problem with a binomial distribution, with p = 0.25 and n = 12. The expected value of X is simply n times p, so 12 * 0.25 = 3 The standard deviation is expressed as d = sqrt(np(1-p)) where d = standard deviation n = number of selections p = probability of selection Substituting known values, we get d = sqrt(12 * 0.25(1-0.25)) d = sqrt(12 * 0.25 * 0.75) d = sqrt(2.25) d = 1.5 Word of warning, the formula doesn't closely approximate a standard deviation curve unless the following is true. n > 9(1-p)/p and n > 9p/(1-p) With the p value of 0.25, the limits are 9(1-0.25)/0.25 = 9*0.75/0.25 = 27 and 9*0.25/(1-0.25) = 9*0.25/0.75 = 3 So n should be at least 28 and 12 is rather small for a proper distribution.

That’s what i got when i looked up what the solutions were

Heptagon is the correct ans