really hard probablity problem

**cubs3205** · October 30th 2007, 04:55 PM

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?

**Plato** · October 30th 2007, 05:06 PM

$\begin{array}{l}<br /> P(R \cap H) = \frac{1}{6}\,,\,P(R \cup H) = \frac{1}{3}\,\& \,P(R) = \frac{3}{{10}} \\ <br /> P(R \cup H) = P(R) + P(H) - P(R \cap H) \\ <br /> \end{array}$

**Soroban** · October 30th 2007, 05:56 PM

Hello, cubs3205!

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?

If you're doing Probability problems, you should know this formula:

. . . . . $P(A \cup B) \;=\;P(A) + P(B) - P(A \cap B)$

We are given: . $\begin{array}{ccc} P(\text{Rain}) & = & \frac{3}{10} \\<br /> P(\text{Rain} \cap \text{Hail}) & = & \frac{1}{6} \\ P(\text{Rain} \cup \text{Hail}) & = & \frac{1}{3}\end{array}$

We have: . $\underbrace{P(\text{Rain} \cup \text{Hail})} \;=\;\underbrace{P(\text{Rain})}\: +\: P(\text{Hail}) - \underbrace{P(\text{Rain} \cap \text{Hail})}$

. . . . . . . . . . . . $\frac{1}{3}\qquad\quad\; =\quad\;\; \frac{3}{10}\quad\;\;+\: P(\text{Hail})\quad\; - \quad \frac{1}{6}$

Therefore: . $P(\text{Hail}) \;=\;\frac{1}{3} - \frac{3}{10} + \frac{1}{6} \;=\;\boxed{\frac{1}{5}}$

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