[amp_mcq option1=”$$\\frac{1}{7}$$” option2=”$$\\frac{9}{{49}}$$” option3=”$$\\frac{{12}}{{49}}$$” option4=”$$\\frac{3}{7}$$” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\frac{12}{49}}$.<\/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 first ball drawn is red” and event B is “the second ball drawn is blue”. We are asked to find the probability of event A happening, given that event B has already happened.<\/p>\n
The probability of event A happening is the number of ways to draw a red ball from the first container and a blue ball from the second container, divided by the total number of ways to draw two balls from the two containers. This is:<\/p>\n
$$P(A) = \\frac{4 \\times 3}{{7 \\times 8}} = \\frac{3}{28}$$<\/p>\n
The probability of event B happening is the number of ways to draw a blue ball from the second container, divided by the total number of ways to draw a ball from the second container. This is:<\/p>\n
$$P(B) = \\frac{3}{{7}}$$<\/p>\n
Therefore, the probability of event A happening, given that event B has already happened, is:<\/p>\n
$$P(A|B) = \\frac{P(A \\cap B)}{P(B)} = \\frac{\\frac{4 \\times 3}{{7 \\times 8}} \\times \\frac{3}{{7}}}{\\frac{3}{{7}}} = \\frac{12}{49}$$<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$\\frac{1}{7}$$” option2=”$$\\frac{9}{{49}}$$” option3=”$$\\frac{{12}}{{49}}$$” option4=”$$\\frac{3}{7}$$” 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-20315","post","type-post","status-publish","format-standard","hentry","category-probability-and-statistics","no-featured-image-padding"],"yoast_head":"\n