Conditional probability is a measure of the probability of an event given that another event has occurred. If the event of interest is A and the event B is known or assumed to have occurred, "the conditional probability of A given B", or "the probability of A under the condition B", is usually written as P(A|B), or sometimes
![P_B(A)](https://tex.z-dn.net/?f=P_B%28A%29)
.
The conditional probability of event A happening, given that event B has happened, written as P(A|B) is given by
![P(A|B)= \frac{P(A \cap B)}{P(B)}](https://tex.z-dn.net/?f=P%28A%7CB%29%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28B%29%7D%20)
In the question, we were told that there are three randomly selected coins which can be a nickel, a dime or a quarter.
The probability of selecting one coin is
![\frac{1}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B3%7D%20)
Part A:
To find <span>the probability that all three coins are quarters if the first two envelopes Jeanne opens each contain a quarter, let the event that all three coins are quarters be A and the event that the first two envelopes Jeanne opens each contain a quarter be B.
P(A) means that the first envelope contains a quarter AND the second envelope contains a quarter AND the third envelope contains a quarter.
Thus
![P(A)= \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}](https://tex.z-dn.net/?f=P%28A%29%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B3%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B3%7D%20%3D%20%5Cfrac%7B1%7D%7B27%7D%20)
</span><span>P(B) means that the first envelope contains a quarter AND the
second envelope contains a quarter
</span><span>Thus
![P(B)= \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}](https://tex.z-dn.net/?f=P%28B%29%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B3%7D%20%3D%20%5Cfrac%7B1%7D%7B9%7D%20)
Therefore,
![P(A|B)=\left( \frac{ \frac{1}{27} }{ \frac{1}{9} } \right)= \frac{1}{3}](https://tex.z-dn.net/?f=P%28A%7CB%29%3D%5Cleft%28%20%5Cfrac%7B%20%5Cfrac%7B1%7D%7B27%7D%20%7D%7B%20%5Cfrac%7B1%7D%7B9%7D%20%7D%20%5Cright%29%3D%20%5Cfrac%7B1%7D%7B3%7D%20)
Part B:
</span>To find the probability that all three coins are different if the first envelope Jeanne opens contains a dime<span>, let the event that all three coins are different be C and the event that the first envelope Jeanne opens contains a dime be D.
</span><span>
![P(C)= \frac{3}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{6}{27} = \frac{2}{9}](https://tex.z-dn.net/?f=P%28C%29%3D%20%5Cfrac%7B3%7D%7B3%7D%20%5Ctimes%20%5Cfrac%7B2%7D%7B3%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B3%7D%20%3D%20%5Cfrac%7B6%7D%7B27%7D%20%3D%20%5Cfrac%7B2%7D%7B9%7D%20)
</span><span>
![P(D)= \frac{1}{3}](https://tex.z-dn.net/?f=P%28D%29%3D%20%5Cfrac%7B1%7D%7B3%7D%20)
</span><span>
Therefore,
![P(C|D)=\left( \frac{ \frac{2}{9} }{ \frac{1}{3} } \right)= \frac{2}{3}](https://tex.z-dn.net/?f=P%28C%7CD%29%3D%5Cleft%28%20%5Cfrac%7B%20%5Cfrac%7B2%7D%7B9%7D%20%7D%7B%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%5Cright%29%3D%20%5Cfrac%7B2%7D%7B3%7D%20)
</span>