[amp_mcq option1=”$$\\frac{4}{{15}}$$” option2=”$$\\frac{7}{{16}}$$” option3=”$$\\frac{1}{2}$$” option4=”$$\\frac{8}{{15}}$$” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\frac{7}{16}}$.<\/p>\n
The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is drawing a red ball and leaving it out, and event B is drawing a green ball.<\/p>\n
The probability of drawing a red ball is $\\frac{8}{16}$, because there are 8 red balls and 16 total balls.<\/p>\n
The probability of drawing a green ball after drawing a red ball is $\\frac{8}{15}$, because there are 8 green balls left and 15 total balls left.<\/p>\n
Therefore, the probability of drawing a red ball, then a green ball is $\\frac{8}{16} \\cdot \\frac{8}{15} = \\boxed{\\frac{7}{16}}$.<\/p>\n
Option A is incorrect because it is the probability of drawing a green ball, then a red ball. This is not the same as the probability of drawing a red ball, then a green ball.<\/p>\n
Option B is incorrect because it is the probability of drawing a red ball on the first draw. This is not the same as the probability of drawing a red ball, then a green ball.<\/p>\n
Option C is incorrect because it is the probability of drawing a green ball on the first draw, then a red ball on the second draw. This is not the same as the probability of drawing a red ball, then a green ball.<\/p>\n
Option D is incorrect because it is the probability of drawing a red ball or a green ball on the first draw, then a red ball or a green ball on the second draw. This is not the same as the probability of drawing a red ball, then a green ball.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$\\frac{4}{{15}}$$” option2=”$$\\frac{7}{{16}}$$” option3=”$$\\frac{1}{2}$$” option4=”$$\\frac{8}{{15}}$$” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[691],"tags":[],"class_list":["post-20329","post","type-post","status-publish","format-standard","hentry","category-probability-and-statistics","no-featured-image-padding"],"yoast_head":"\n