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zhuklara [117]
3 years ago
7

In a binomial experiment, the probability of success is .06. What is the probability of two successes in seven trials? a. .0554

b. .28 c. .0036 d. .06
Mathematics
1 answer:
zaharov [31]3 years ago
4 0

Answer:

a. .0554

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The probability of success is .06.

This means that p = 0.06

What is the probability of two successes in seven trials?

This is P(X = 2) when n = 7. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{7,2}.(0.06)^{2}.(0.94)^{5} = 0.054

The correct answer is given by option a.

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7+6 (mod 5) =<br><br>2+1 (mod 5) =<br><br>20 (mod 11) =<br><br>35 (mod 11) =<br><br>20 (mod 11) =​
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======================================

Explanation:

7+6 = 13. Divide this over 5 to get 13/5 = 2 remainder 3. The quotient 2 is something we don't care about. <u>We only worry about the remainder</u>. Therefore 7+6 = 3 (mod 5)

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2+1 = 3 (mod 5) for similar reasoning as above. 3/5 = 0 remainder 3.

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20 = 9 (mod 11) since 20/11 = 1 remainder 9. Imagine you had 20 cookies and 11 friends. Each friend would get 1 whole cookie (quotient) and there could be 9 left over (remainder).

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Or you could use repeated subtraction like so to find the remainder

35-11 = 24

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13-11 = 2

The last result (2) is smaller than 11, so we stop here and this is the remainder.

Therefore, 35 = 2 (mod 11)

--------------

Here's another way to think of it. Consider you have $35 in your pocket. Let's say a store is selling trinkets for $11 each. We can pose these key questions:

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The answer to the first question is 3 trinkets because 3*11 = 33 dollars is under the budget of $35. The amount left over is 35-33 = 2 dollars which is the remainder. It's not larger than 11, so we cannot buy any more trinkets at this point.

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Answer:

Step-by-step explanation:

\large\underline{\sf{Solution-}}

<h2 /><h2><u>Consider</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)

<h2><u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>

\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx

\rm \: {cos \: (2\pi + x) }

\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanx

\rm \: cot(2\pi + x) \: = \: cotx

So, on substituting all these values, we get

\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}

\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}

\rm \: = \: 1=1

<h2>Hence,</h2>

\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

  • sin (90°-θ)  =  cos θ
  • cos (90°-θ)  =  sin θ
  • tan (90°-θ)  =  cot θ
  • csc (90°-θ)  =  sec θ
  • sec (90°-θ)  =  csc θ
  • cot (90°-θ)  =  tan θ
  • sin (90°+θ)  =  cos θ
  • cos (90°+θ)  =  -sin θ
  • tan (90°+θ)  =  -cot θ
  • csc (90°+θ)  =  sec θ
  • sec (90°+θ)  =  -csc θ
  • cot (90°+θ)  =  -tan θ
  • sin (180°-θ)  =  sin θ
  • cos (180°-θ)  =  -cos θ
  • tan (180°-θ)  =  -tan θ
  • csc (180°-θ)  =  csc θ
  • sec (180°-θ)  =  -sec θ
  • cot (180°-θ)  =  -cot θ
  • sin (180°+θ)  =  -sin θ
  • cos (180°+θ)  =  -cos θ
  • tan (180°+θ)  =  tan θ
  • csc (180°+θ)  =  -csc θ
  • sec (180°+θ)  =  -sec θ
  • cot (180°+θ)  =  cot θ
  • sin (270°-θ)  =  -cos θ
  • cos (270°-θ)  =  -sin θ
  • tan (270°-θ)  =  cot θ
  • csc (270°-θ)  =  -sec θ
  • sec (270°-θ)  =  -csc θ
  • cot (270°-θ)  =  tan θ
  • sin (270°+θ)  =  -cos θ
  • cos (270°+θ)  =  sin θ
  • tan (270°+θ)  =  -cot θ
  • csc (270°+θ)  =  -sec θ
  • sec (270°+θ)  =  cos θ
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