Suppose S and T are mutually exclusive events. Find P(S or T) if P(S) = 1/3 and P(T) = 5/12?

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Suppose S and T are mutually exclusive events. Find P(S or T) if P(S) = 1/3 and P(T) = 5/12?

3/4
7/18
5/36

2 Answers

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Answer 1 · 2017-04-20T13:59:12

In an OR situation you may ADD

Explanation:

$P (S or T) = P (S) + P (T) = \frac{1}{3} + \frac{5}{12} = \frac{4}{12} + \frac{5}{12} = \frac{9}{12} = \frac{3}{4}$ $P(SorT)=P(S)+P(T)=1/3+5/12=4/12+5/12=9/12=3/4$

Answer 2 · 2017-04-24T11:47:41

$P (S \cup T) = \frac{3}{4}$ $P(S uu T) = 3/4$

Explanation:

We use the following fundamental definition from Probability and Set Theory:

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$ $P(A uu B) = P(A) + P(B) - P(A nn B)$

And if we apply this to our problem we have:

$P (S \cup T) = P (S) + P (T) - P (S \cap T)$ $P(S uu T) = P(S) + P(T) - P(S nn T)$
$= \frac{1}{3} + \frac{5}{12} - P (S \cap T)$ $" " = 1/3 + 5/12 - P(S nn T)$
$= \frac{3}{4} - P (S \cap T)$ $" " = 3/4 - P(S nn T)$

Now we are told that $S$ $S$ and $T$ $T$ are mutually exclusive, and so:

$P (S \cap T) = 0$ $P(S nn T) =0$

Hence,

$P (S \cup T) = \frac{3}{4}$ $P(S uu T) = 3/4$

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Suppose S and T are mutually exclusive events. Find P(S or T) if P(S) = 1/3 and P(T) = 5/12?

3/4
7/18
5/36

2 Answers

Explanation:

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

Suppose S and T are mutually exclusive events. Find P(S or T) if P(S) = 1/3 and P(T) = 5/12?

3/4 7/18 5/36

2 Answers

Explanation:

Explanation:

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Impact of this question

3/4
7/18
5/36