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CaHeK987 [17]
3 years ago
12

Kevin installed a certain brand of automatic garage door opener that utilizes a transmitter control with four independent​ switc

hes, each one set on or off. The receiver​ (wired to the​ door) must be set with the same pattern as the transmitter. If six neighbors with the same type of opener set their switches​ independently, what is the probability of at least one pair of neighbors using the same​ settings?
Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
3 0

Kevin installed a certain brand of automatic garage door opener that utilizes a transmitter control with four independent​ switches, each one set on or off. The receiver​ (wired to the​ door) must be set with the same pattern as the transmitter. If six neighbors with the same type of opener set their switches​ independently.<u>The probability of at least one pair of neighbors using the same​ settings is 0.65633</u>

Step-by-step explanation:

<u>Step 1</u>

In the question it is given that

Automatic garage door opener utilizes a transmitter control with four independent​ switches

<u>So .the number of Combinations possible with the Transmitters </u>=

2*2*2*2= 16

<u> Step 2</u>

Probability of at least one pair of neighbors using the same settings = 1- Probability of All Neighbors using different settings.

= 1- 16*15*14*13*12*11/(16^6)

<u> Step 3</u>

Probability of at least one pair of neighbors using the same settings=

= 1- 0.343666

<u> Step 4</u>

<u>So the probability of at least </u>one pair of neighbors using the same settings

is  0.65633

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\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\  \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}

To get cos 240 degrees:

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In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

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\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\  \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}

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