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Math Help - really hard probablity problem

  1. #1
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    really hard probablity problem

    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?
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  2. #2
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    \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}
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  3. #3
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    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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