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suter [353]
3 years ago
11

If two events are mutually exclusive, what is the probability that one or the other occurs?

Mathematics
1 answer:
tekilochka [14]3 years ago
8 0
Mutually Exclusive Events
Another word that means mutually exclusive is disjoint. If two events are disjoint, then the probability of them both occurring at the same time is 0. If two events are mutually exclusive, then the probability of either occurring is the sum of the probabilities of each occurring.
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Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0
andreev551 [17]

Recall the angle sum identity for cosine:

cos(<em>x</em> + <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) - sin(<em>x</em>) sin(<em>y</em>)

cos(<em>x</em> - <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) + sin(<em>x</em>) sin(<em>y</em>)

==>   sin(<em>x</em>) sin(<em>y</em>) = 1/2 (cos(<em>x</em> - <em>y</em>) - cos(<em>x</em> + <em>y</em>))

Then rewrite the equation as

sin(4<em>x</em>) sin(5<em>x</em>) + sin(4<em>x</em>) sin(3<em>x</em>) - sin(2<em>x</em>) sin(<em>x</em>) = 0

1/2 (cos(-<em>x</em>) - cos(9<em>x</em>)) + 1/2 (cos(<em>x</em>) - cos(7<em>x</em>)) - 1/2 (cos(<em>x</em>) - cos(3<em>x</em>)) = 0

1/2 (cos(9<em>x</em>) - cos(<em>x</em>)) + 1/2 (cos(7<em>x</em>) - cos(3<em>x</em>)) = 0

sin(5<em>x</em>) sin(-4<em>x</em>) + sin(5<em>x</em>) sin(-2<em>x</em>) = 0

-sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

Recall the double angle identity for sine:

sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)

Rewrite the equation again as

sin(5<em>x</em>) (2 sin(2<em>x</em>) cos(2<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) sin(2<em>x</em>) (2 cos(2<em>x</em>) + 1) = 0

sin(5<em>x</em>) = 0   <u>or</u>   sin(2<em>x</em>) = 0   <u>or</u>   2 cos(2<em>x</em>) + 1 = 0

sin(5<em>x</em>) = 0   <u>or</u>   sin(2<em>x</em>) = 0   <u>or</u>   cos(2<em>x</em>) = -1/2

sin(5<em>x</em>) = 0   ==>   5<em>x</em> = arcsin(0) + 2<em>nπ</em>   <u>or</u>   5<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … …   ==>   5<em>x</em> = 2<em>nπ</em>   <u>or</u>   5<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … …   ==>   <em>x</em> = 2<em>nπ</em>/5   <u>or</u>   <em>x</em> = (2<em>n</em> + 1)<em>π</em>/5

sin(2<em>x</em>) = 0   ==>   2<em>x</em> = arcsin(0) + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … …   ==>   2<em>x</em> = 2<em>nπ</em>   <u>or</u>   2<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … …   ==>   <em>x</em> = <em>nπ</em>   <u>or</u>   <em>x</em> = (2<em>n</em> + 1)<em>π</em>/2

cos(2<em>x</em>) = -1/2   ==>   2<em>x</em> = arccos(-1/2) + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = -arccos(-1/2) + 2<em>nπ</em>

… … … … … …    ==>   2<em>x</em> = 2<em>π</em>/3 + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = -2<em>π</em>/3 + 2<em>nπ</em>

… … … … … …    ==>   <em>x</em> = <em>π</em>/3 + <em>nπ</em>   <u>or</u>   <em>x</em> = -<em>π</em>/3 + <em>nπ</em>

<em />

(where <em>n</em> is any integer)

5 0
3 years ago
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