[amp_mcq option1=”$$\\frac{1}{4},\\,\\frac{1}{2}$$” option2=”$$\\frac{1}{2},\\,\\frac{1}{4}$$” option3=”$$\\frac{1}{2}$$, 1″ option4=”1, $$\\frac{1}{2}$$” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{C. }\\frac{1}{2}, 1}$.<\/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 is calculated as follows:<\/p>\n
$$P(A|B) = \\frac{P(A \\cap B)}{P(B)}$$<\/p>\n
where $P(A \\cap B)$ is the probability of both events A and B happening.<\/p>\n
In this case, we are given that $P(A) = 1$ and $P(B) = \\frac{1}{2}$. We can then calculate $P(A \\cap B)$ as follows:<\/p>\n
$$P(A \\cap B) = P(A) \\cdot P(B) = 1 \\cdot \\frac{1}{2} = \\frac{1}{2}$$<\/p>\n
Therefore, the conditional probability of A given B is:<\/p>\n
$$P(A|B) = \\frac{P(A \\cap B)}{P(B)} = \\frac{\\frac{1}{2}}{\\frac{1}{2}} = 1$$<\/p>\n
The probability of event B happening, given that event A has already happened, is called the conditional probability of B given A, and is denoted by $P(B|A)$. It is calculated as follows:<\/p>\n
$$P(B|A) = \\frac{P(A \\cap B)}{P(A)}$$<\/p>\n
where $P(A \\cap B)$ is the probability of both events A and B happening.<\/p>\n
In this case, we are given that $P(A) = 1$ and $P(B) = \\frac{1}{2}$. We can then calculate $P(A \\cap B)$ as follows:<\/p>\n
$$P(A \\cap B) = P(A) \\cdot P(B) = 1 \\cdot \\frac{1}{2} = \\frac{1}{2}$$<\/p>\n
Therefore, the conditional probability of B given A is:<\/p>\n
$$P(B|A) = \\frac{P(A \\cap B)}{P(A)} = \\frac{\\frac{1}{2}}{1} = \\frac{1}{2}$$<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$\\frac{1}{4},\\,\\frac{1}{2}$$” option2=”$$\\frac{1}{2},\\,\\frac{1}{4}$$” option3=”$$\\frac{1}{2}$$, 1″ option4=”1, $$\\frac{1}{2}$$” 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-20322","post","type-post","status-publish","format-standard","hentry","category-probability-and-statistics","no-featured-image-padding"],"yoast_head":"\n