What is Conditional Probability?

Solution

• Part A: The probability that a randomly selected plant grew successfully is calculated by dividing the number of plants that grew successfully by the total number of plants: 
  \[ \begin{align*}P(\text{Grew Successfully}) &= \frac{n(\text{Grew Successfully})}{n(\text{{Total}})}\\&= \frac{{125}}{{200}}\\\\ &= 0.625\end{align*} \]
  Thus, the probability that a randomly selected plant grew successfully is 0.625.
• Part B: The probability that a plant grew successfully given that it received fertilizer is found by focusing only on plants that received fertilizer.
  

  Plant Growth and Fertilizer Use

  Plant Growth	Fertilizer Used
  Grew Successfully	85
  Did Not Grow	30
  Total	115

  Notice that the sample space has been redefined from the original 200 plants to the 115 plants where fertilizer was used.  In terms of the table, we ignore all the columns and just focus on the column "Fertilier Used."  We compute the probability as normal with this new sample space, but notice that the notation used still refers to the original table:
  \[\begin{align*} P(\text{Grew Successfully if Fertilizer Used}) &= \frac{n(\text{Grew Successfully and Fertilizer Used})}{n(\text{Fertilizer Used})}\\\\&= \frac{{85}}{{115}}\\\\& \approx 0.739\end{align*} \]
  Thus, the probability that a plant grew successfully given that it received fertilizer is 0.739.

$$\tag*{\(\blacksquare\)}$$

Solution

• Part A: We first have to make two preliminary calculations: \(P(\text{{Moth}})\) and \(P(\text{Nocturnal and Moth})\):
  \[ \begin{align*} P(\text{{Moth}}) & = \frac{n(\text{{Moth}})}{n(\text{{Total}})} \\\\ & = \frac{{150}}{{250}} \\\\ & = 0.6 \end{align*} \]
  \[ \begin{align*} P(\text{Nocturnal and Moth})&=\dfrac{n(\text{Nocturnal and Moth})}{\text{{Total}}} \\\\&=\dfrac{{120}}{{250}}\\\\&=0.48\end{align*}\]
  Now, we can use the definition to compute the conditional probabiltiy:
  \[\begin{align*}P(\text{{Nocturnal}}\mid\text{{Moth}})&=\dfrac{P(\text{Nocturnal and Moth})}{P(\text{{Moth}})}\\\\&=\dfrac{0.48}{0.60}\\\\&=0.8\end{align*}\]
  Thus, the probability that a randomly selected moth is nocturnal is 0.8.
• Part B: The probability that a randomly selected insect is nocturnal, given that it is a moth, is found by focusing only on moths.
  

  Insect Species and Nocturnal Behavior

  Insect Type	Nocturnal	Diurnal	Total
  Moth	120	30	150

  Notice that the sample space has been redefined from the original 250 insects to the 150 moths observed. In terms of the table, we ignore all rows except for "Moth." We compute the probability as normal with this new sample space, but the notation still refers to the original table:
  \[ \begin{align*} P(\text{{Nocturnal}} \mid \text{{Moth}}) & = \frac{n(\text{Nocturnal and Moth})}{n(\text{{Moth}})} \\\\ & = \frac{{120}}{{150}} \\\\ & = 0.8 \end{align*} \]
  Thus, the probability that an insect is nocturnal given that it is a moth is 0.8.

$$\tag*{\(\blacksquare\)}$$

Solution

After picking the initial white sock, there are now 5 white socks remaining, the 4 initial black socks, and the remain sock total is now reduced to 9.  The probability of picking a white sock as the second sock, given that the first sock was also white, is calculated the same way as any other probability; we simply use the updated numbers instead of the original ones. 
\[ \begin{align*} P(\text{Second Sock White} \mid \text{First Sock White}) & = \frac{n(\text{Remaining White Socks})}{n(\text{Remaining Total Socks})} \\\\ & = \frac{{5}}{{9}} \\\\ & \approx 0.556 \end{align*} \]
Thus, the probability that the second sock picked is also white, given that the first sock picked was white, is 0.556.

$$\tag*{\(\blacksquare\)}$$