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Ganezh [65]
3 years ago
7

A drawer contains 6 black neckties, 2 white neckties,4 red neckties,2 maroon neckties and 2 blue neckties. One necktie is picked

at random and then replaced. This is repeated 300 times. Predict how many times you can expect the color of the necktie to be: A. White B. Blue C. Red D. Black E. Maroon F. Not white
Mathematics
1 answer:
kogti [31]3 years ago
4 0

A) The probability of picking a white tie 300 times = (\frac{1}{8}) ^{300}

B)  The probability of picking a blue tie 300 times = (\frac{1}{8}) ^{300}

C) The probability of picking a red tie 300 times = (\frac{1}{4}) ^{300}

D) the probability of picking a black tie 300 times = (\frac{3}{8}) ^{300}

E ) the probability of picking a maroon tie 300 times = (\frac{1}{8}) ^{300}

F) the probability of NOT picking a white tie 300 times = (\frac{7}{8}) ^{300}

Step-by-step explanation:

Here, the total number of black neckties  = 6

The total number of white neckties  = 2

The total number of red neckties  = 4

The total number of maroon neckties  = 2

The total number of blue neckties  = 2

The number of times the experiment is repeated = 300

A )  P(Picking a white tie)  = \frac{\textrm{Total number of white ties}}{\textrm{Total Bow ties}}

= \frac{2}{16}  = \frac{1}{8}

So, the probability of picking a white ONCE is 1/8.

Now, as the experiment is REPEATED 300 times with replacement.

So, the probability of picking a white tie 300 times = (\frac{1}{8}) ^{300}

B) P(Picking a BLUE  tie)  = \frac{\textrm{Total number of blue ties}}{\textrm{Total Bow ties}} = \frac{2}{16}  = \frac{1}{8}

So, the probability of picking a blue ONCE is 1/8.

Hence, the probability of picking a blue tie 300 times = (\frac{1}{8}) ^{300}

C) P(Picking a Red  tie)  = \frac{\textrm{Total number of Red ties}}{\textrm{Total Bow ties}} = \frac{4}{16}  = \frac{1}{4}

So, the probability of picking a red ONCE is 1/4.

Hence, the probability of picking a red tie 300 times = (\frac{1}{4}) ^{300}

D) P(Picking a Black  tie)  = \frac{\textrm{Total number of black ties}}{\textrm{Total Bow ties}} = \frac{6}{16}  = \frac{3}{8}

So, the probability of picking a red ONCE is 3/8.

Hence, the probability of picking a black tie 300 times = (\frac{3}{8}) ^{300}

E) P(Picking a maroon  tie)  = \frac{\textrm{Total number of maroon ties}}{\textrm{Total Bow ties}} = \frac{2}{16}  = \frac{1}{8}

So, the probability of picking a maroon ONCE is 1/8.

Hence, the probability of picking a maroon tie 300 times = (\frac{1}{8}) ^{300}

F) P(Picking a NOT whiten  tie)  = 1 - P( picking a white tie)

= 1-(\frac{1}{8} ) = \frac{8-1}{8}  = (\frac{7}{8} )

So, the probability of NOT  picking a white ONCE is 7/8.

Hence, the probability of NOT picking a white tie 300 times = (\frac{7}{8}) ^{300}

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Step-by-step explanation:

To find the point where the functions will intersect, we want to find the place that they are equal to each other. We do this by setting the equations equal to each other.

5x+1=-2x+15                       [add both sides by 2x]

7x+1=15                               [subtract both sides by 1]

7x=14                                  [divide both sides by 7]

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Now, we know that the functions intersect at x=2. To find the point, we would plug x back into both equations to ensure that the points are the same.

f(2)=5(2)+1                         [multiply]

f(2)=10+1                            [add]

f(2)=11

The point on f(x) is (2,11).

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g(2)=-4+15                         [add]

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We have to find if the above sequence is arithmetic sequence or not

An arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant

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<em><u>Let us find the difference between terms</u></em>

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Thus the difference between terms is not constant

So the given sequence is not arithmetic sequence

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