The probablity that it will rain and hail in one week is 1/6. The probablity that there will be rain, hail, or both in one week is 1/3. The probablity that it will rain in one week is 3/10. What is the probablity that it will hail in any given week?
The probablity that it will rain and hail in one week is 1/6. The probablity that there will be rain, hail, or both in one week is 1/3. The probablity that it will rain in one week is 3/10. What is the probablity that it will hail in any given week?
Hello, cubs3205!
If you're doing Probability problems, you should know this formula:The probablity that it will rain and hail in one week is 1/6.
The probablity that there will be rain, hail, or both in one week is 1/3.
The probablity that it will rain in one week is 3/10.
What is the probablity that it will hail in any given week?
. . . . . $\displaystyle P(A \cup B) \;=\;P(A) + P(B) - P(A \cap B)$
We are given: .$\displaystyle \begin{array}{ccc} P(\text{Rain}) & = & \frac{3}{10} \\
P(\text{Rain} \cap \text{Hail}) & = & \frac{1}{6} \\ P(\text{Rain} \cup \text{Hail}) & = & \frac{1}{3}\end{array}$
We have: .$\displaystyle \underbrace{P(\text{Rain} \cup \text{Hail})} \;=\;\underbrace{P(\text{Rain})}\: +\: P(\text{Hail}) - \underbrace{P(\text{Rain} \cap \text{Hail})}$
. . . . . . . . . . . .$\displaystyle \frac{1}{3}\qquad\quad\; =\quad\;\; \frac{3}{10}\quad\;\;+\: P(\text{Hail})\quad\; - \quad \frac{1}{6}$
Therefore: .$\displaystyle P(\text{Hail}) \;=\;\frac{1}{3} - \frac{3}{10} + \frac{1}{6} \;=\;\boxed{\frac{1}{5}}$