[amp_mcq option1=”$$\\frac{1}{3}$$” option2=”$$\\frac{3}{7}$$” option3=”$$\\frac{1}{2}$$” option4=”$$\\frac{4}{7}$$” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\frac{3}{7}}$.<\/p>\n
The probability of event A happening, given that event B has already happened, is called the conditional probability of A given B, and is denoted by $P(A|B)$. It can be calculated using the following formula:<\/p>\n
$$P(A|B) = \\frac{P(A \\cap B)}{P(B)}$$<\/p>\n
In this case, event A is the event that the second ball removed is red, and event B is the event that the first ball removed is white. We are given that event B has already happened, so we need to calculate $P(A \\cap B)$ and $P(B)$.<\/p>\n
$P(A \\cap B)$ is the probability that both the first ball removed is white and the second ball removed is red. This can be calculated as follows:<\/p>\n
$$P(A \\cap B) = \\frac{4}{7} \\times \\frac{3}{6} = \\frac{2}{7}$$<\/p>\n
$P(B)$ is the probability that the first ball removed is white. This can be calculated as follows:<\/p>\n
$$P(B) = \\frac{4}{7}$$<\/p>\n
Therefore, the conditional probability that the second ball removed is red, given that the first ball removed is white, is:<\/p>\n
$$P(A|B) = \\frac{P(A \\cap B)}{P(B)} = \\frac{\\frac{2}{7}}{\\frac{4}{7}} = \\frac{3}{7}$$<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$\\frac{1}{3}$$” option2=”$$\\frac{3}{7}$$” option3=”$$\\frac{1}{2}$$” option4=”$$\\frac{4}{7}$$” 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-20333","post","type-post","status-publish","format-standard","hentry","category-probability-and-statistics","no-featured-image-padding"],"yoast_head":"\n