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.