An urn is filled with black balls and white balls only. There are 45 balls in total. If the probability of randomly drawing a white ball is 4/5, how many white balls must be added to the urn so that the probability of randomly drawing a white ball is 7/8? To prevent "plugging and chugging," I'll leave this as an open-ended question.

Brent Hanneson wrote:An urn is filled with black balls and white balls only. There are 45 balls in total. If the probability of randomly drawing a white ball is 4/5, how many white balls must be added to the urn so that the probability of randomly drawing a white ball is 7/8? To prevent "plugging and chugging," I'll leave this as an open-ended question. White balls = W W/45 = 4/5 W = 36 Let x be the number of white balls to be added 36+x / 45 + x = 7/8 288 + 8x = 315 + 7x x = 27 Hence, 27 balls must be added. OA?

B and WHite balss B+W=45 4/5=F/45 36 white balls .. and 9 balck balls so the probality !! 7/8=36+x/45+x solve for x 27 balss must be added

An urn is filled with black balls and white balls only. There are 45 balls in total. If the probability of randomly drawing a white ball is 4/5, how many white balls must be added to the urn so that the probability of randomly drawing a white ball is 7/8? x + y = 45 x/ x+y = 4/5 x = 36 y = 9 7/8 = 36+k / 45+ k k = 27 hence 27 balls have to be added

For the sake of simplified calculation, we can set up the ratio for white ball to black ball - (36+x)/9 = 7/1

Out of curiosity, what level do you all think this is?

txeconomist wrote:Out of curiosity, what level do you all think this is? On a scale of 1 to 10 with 10 being the hardest level, I'd say it's 7. But I'm not an expert.