[amp_mcq option1=”$$\\frac{{10}}{{21}}$$” option2=”$$\\frac{5}{{12}}$$” option3=”$$\\frac{2}{3}$$” option4=”$$\\frac{1}{6}$$” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\frac{10}{21}}$.<\/p>\n
The probability of getting at least 6 on two rolls of a six-sided die is the sum of the probabilities of getting 6 on the first roll and 5 or more on the second roll, getting 5 on the first roll and 6 or more on the second roll, and getting 4 on the first roll and 7 or more on the second roll.<\/p>\n
The probability of getting 6 on the first roll is $\\frac{1}{6}$. The probability of getting 5 or more on the second roll is $\\frac{5}{6}$, since there are 5 possible outcomes that satisfy this condition (namely, 5, 6, 7, 8, and 9). The probability of getting 5 on the first roll and 6 or more on the second roll is $\\frac{1}{6} \\cdot \\frac{5}{6} = \\frac{5}{36}$. The probability of getting 4 on the first roll and 7 or more on the second roll is $\\frac{1}{6} \\cdot \\frac{5}{6} = \\frac{5}{36}$.<\/p>\n
Therefore, the probability of getting at least 6 on two rolls of a six-sided die is $\\frac{1}{6} + \\frac{5}{36} + \\frac{5}{36} = \\boxed{\\frac{10}{21}}$.<\/p>\n
Option A is incorrect because it is the probability of getting at least 6 on a single roll of a six-sided die. Option B is incorrect because it is the probability of getting 5 or more on a single roll of a six-sided die. Option C is incorrect because it is the probability of getting 2 or more on a single roll of a six-sided die. Option D is incorrect because it is the probability of getting 1 on a single roll of a six-sided die.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$\\frac{{10}}{{21}}$$” option2=”$$\\frac{5}{{12}}$$” option3=”$$\\frac{2}{3}$$” option4=”$$\\frac{1}{6}$$” 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-20302","post","type-post","status-publish","format-standard","hentry","category-probability-and-statistics","no-featured-image-padding"],"yoast_head":"\n