Given Probability

<a href="https://www.prepswift.com/quizzes/quiz/prepswift-given-probability" target="_blank">Given Probability Exercise</a>Given Probability can be a tricky beast. In a nutshell, a given probability calculation assumes that only SOME cases are valid (not all of them). This changes the value of the denominator.  An Easy Contrast to Illustrate Given Probability Scenario 1: An alien flips a fair coin twice. What is the probability it gets heads twice? $$\left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4}$$ Scenario 2: An alien flips a fair coin twice. Given that the first flip was heads, what is the probability that the second flip will be heads? Well, that's just $\frac{1}{2}$ right? Yes, that is correct. Notice how the denominator in the first scenario is $4$. This makes sense because we're looking at four different possibilities, with only one of the possibilities making us "happy."  TT, HT, TH, HH But in the second scenario, we're actually excluding two of the cases because we're "assuming" that the first flip must be heads. It is "given" that the first flip is heads. So the cases where tails is first are out. <s>TT</s>, HT, <s>TH</s>, HH Another Example A couple has four children. If it is assumed or "given" that at least two of the children are boys, what is the probability that all four are boys? If we wanted to solve this without the "given" part, it'd be super straightforward. We would simply do $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$. But that won't work here. We have to exclude the cases that have zero boys or only one boy. <ul style="margin-left: 40px;"> <li>$0$ boys: GGGG <ul> <li>How many ways can this happen? </li> <li>$\frac{4!}{4!} = 1$ </li> <li>exclude $1$ case</li> </ul> </li> <li>$1$ boy: BGGG <ul> <li>How many ways can this happen?</li> <li>$\frac{4!}{1!3!} = 4$</li> <li>exclude $4$ cases</li> </ul> </li> <li>$2$ boys: BBGG <ul> <li>How many ways can this happen?</li> <li>$\frac{4!}{2!2!} = 6$</li> <li>$6$ valid cases</li> </ul> </li> <li>$3$ boys: BBBG <ul> <li>How many ways can this happen?</li> <li>$\frac{4!}{3!1!} = 4$</li> <li>$4$ valid cases</li> </ul> </li> <li>$4$ boys: BBBB <ul> <li>How many ways can this happen?</li> <li>$\frac{4!}{4!} = 1$</li> <li>$1$ valid case</li> </ul> </li> </ul> In total, we have $16$ cases but $5$ of them are invalid and must be exluded. So really, we're only dealing with $11$ valid cases. That is the denominator of our probability calculation. How many of those valid cases make us happy? Only $1$, the case in which all four children are boys. $$\frac{1}{11}$$ Final Answer