Math Question: A bag contains 5 red counters, 12 yellow and 8 green. A counter is selected from the bag at random then replaced, and this is repeated 20 times. The random variable X represents the number of times a yellow counter is removed. a Write the distribution of X. b Calculate the probability of removing a yellow counter half of the time. a X ~ _(_,_) b P(X=_)=_

It sounds like they're asking for the PMF (I hope they aren't asking you to draw it by hand). The formula is (n choose k) p^k (1-p)^(n-k). So for this problem n=20 and p=0.48. (Maybe they mean n=21, since counter is selected and replaced and "this is repeated 20 times" which is 1 initial time plus 21 repetitions, but I doubt it.) p=0.48 because 12/(5+12+8)=12/25=(4*12)%=48%=0.48 a) X ~ B(20,0.48) Which sounds more likely to fit in the blanks than the verbose: Pr(X=k)=(20 choose k)(.48)^k *(.52)^(20-k) b) if they want to know the probability of drawing a yellow counter PRECISELY half the time, use the formula above: Pr(X=k)=(20 choose k)(.48)^k *(.52)^(20-k) 10 is half 20, so k=10: Pr(X=10)=(20 choose 10)(.48)^10 *(.52)^(20-10)= 184756(0.00064925062)*(.52)^(20-10)= 184756(0.00064925062)*(.52)^(10)= 184756(0.00064925062)*(.52)^(10)= 184756*(0.00064925062)*0.00144555105= 0.17339810928 Or 17.34% That's pretty close (and under, as it should be) to the 17.6197% we'd get for precisely 10 heads from 20 tosses of a fair coin (p=q=.50). If they want the probability it's AT LEAST half I'd calculate the CDF (Cumulative distribution function) for k=9, then use the complement; that is, the probably that it happens 10 times or more is equal to 1 minus the probability that is happens 9 times or fewer. I don't think that's what they want because "b P(X=_)=_ " Is an equality, not an inequality, for the X value. I think they want P(X=10)=0.1734 or to however many decimal points they specify. For a X ~ _(_,_) I'm thinking X ~ B(n,p) so X ~ B(20,0.48) or X ~ Bin(20,0.48) But it should match what you're seeing in lectures and the textbook.
Answered On 05/27/2020 14:55