[amp_mcq option1=”$$P\\left( {A\/B} \\right) = \\frac{{P\\left( A \\right) \\cdot P\\left( {B\/A} \\right)}}{{P\\left( A \\right)}}$$” option2=”$$P\\left( {A\/B} \\right) = P\\left( A \\right) \\cdot P\\left( {B\/A} \\right)$$” option3=”$$P\\left( {A\/B} \\right) = P\\left( B \\right) \\cdot P\\left( {A\/B} \\right)$$” option4=”$$P\\left( {A\/B} \\right) = \\frac{{P\\left( A \\right) \\cdot P\\left( {B\/A} \\right)}}{{P\\left( B \\right)}}$$” correct=”option2″]<\/p>\n
The correct answer is:<\/p>\n
$$P\\left( {A\/B} \\right) = \\frac{{P\\left( A \\right) \\cdot P\\left( {B\/A} \\right)}}{{P\\left( B \\right)}}$$<\/p>\n
This is the conditional probability formula. It tells us that the probability of event A happening given that event B has already happened is equal to the probability of both events happening divided by the probability of event B happening.<\/p>\n
In other words, if we know that event B has happened, then the probability of event A happening is increased or decreased depending on how likely event A is to happen given that event B has already happened.<\/p>\n
For example, let’s say that we are flipping a coin. The probability of flipping heads is $\\frac{1}{2}$. Let’s also say that we know that the coin has landed on heads on the last 10 flips. The probability of flipping heads on the next flip is still $\\frac{1}{2}$, but the conditional probability of flipping heads given that the last 10 flips were heads is much higher than $\\frac{1}{2}$. This is because we know that the coin is more likely to land on heads than tails after 10 consecutive heads.<\/p>\n
The conditional probability formula can be used to calculate the probability of any event given that another event has already happened. It is a powerful tool that can be used to make predictions about the future.<\/p>\n
The other options are incorrect because they do not take into account the probability of event B happening. Option A is the probability of event A happening, but it does not take into account the fact that event B has already happened. Option B is the probability of both events happening, but it does not take into account the probability of event B happening. Option C is the probability of event B happening, but it does not take into account the probability of event A happening.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$$P\\left( {A\/B} \\right) = \\frac{{P\\left( A \\right) \\cdot P\\left( {B\/A} \\right)}}{{P\\left( A \\right)}}$$” option2=”$$P\\left( {A\/B} \\right) = P\\left( A \\right) \\cdot P\\left( {B\/A} \\right)$$” option3=”$$P\\left( {A\/B} \\right) = P\\left( B \\right) \\cdot P\\left( {A\/B} \\right)$$” option4=”$$P\\left( {A\/B} \\right) = \\frac{{P\\left( A \\right) \\cdot P\\left( {B\/A} \\right)}}{{P\\left( B \\right)}}$$” correct=”option2″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[701],"tags":[],"class_list":["post-33826","post","type-post","status-publish","format-standard","hentry","category-business-statistics-and-research-methods","no-featured-image-padding"],"yoast_head":"\n