According to government data, the probability that an adult was never in a museum is 15%. in a random survey of 10 adults, what is the probability that two or fewer were never in a museum? round to the nearest thousandth.

Let p be the probability that an adult was never in a museum. Hence p = 0.15. Then q is the probability that an adult was in a museum is 1 - 0.15 = 0.75. We have a binomial expansion where the probability of k success in n trials is given by P_n(k) = (n, k) p^(k) q^(n -k) where (n, k) is the number of ways to select 10 objects from k things. At least two or fewer means we have P_10 (< or equal to 2) So we have P_10 (less than or equal to 2) = P_10 ( 0) + P_10 (1) + P_10 (2). So we have P _10 (0) = ( 10, 0) (0.15)^ (0) (0.75)^(0) = 0.196. For P_10(1), we have 0.3474 and for P_10(2), we have 0.2758. Adding these we have 0.1960 + 0.3474 + 0.2758 = 0.8192.

Best answer that i found on yahoo ok   set :   x= length of 2nd piece in cm  x- 3= length of 1st piece in cm  x-4= length of 3rd piece in cm  then since you know they all came from the 20 cm ribbon you add them together and set them equal to 20x+ x-3 +x-4 = 20  3x - 7= 20  3x= 27  x= 9  the shortest piece is the third piece which is x- 4  x-4  9-4 = 5  the length of the smallest piece is 5 cm.  the other lengths are  9 cm  6 cm