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pickupchik [31]
3 years ago
13

A school class went on a field trip to see a magician perform there were 17 females 20 males in the class the magicians randomly

selected a volunteer from the audience in which had 52 females and 68 males given that the randomly selected audience member is a student from the class which equation can be used to find the probability p that the perdimos also a female?please help does anyone know the answer
Mathematics
1 answer:
zubka84 [21]3 years ago
5 0

Answer:

P(B)= 17/20

Step-by-step explanation:

Hello!

The audience of the magic show is conformed by a total of 120 people, 52 of which are female and 68 are men.

Within the audience there is a school class of 37, of these students, 17 are female and 20 are male.

If a random member of the audience is selected as a volunteer:

Let "A" represent the event that "the selected volunteer is a student of the class"

And "B" the event that "the selected student is female"

You have to calculate the probability of the selected volunteer being female, given that it is a member of the school class.

Symbolically:

P(B|A)

Using the formula of conditional probabilities you can calculate it as:

P(B|A)= \frac{P(AnB)}{P(A)}

P(A∩B)= P(A)*P(B)= (\frac{37}{120} )*(\frac{17}{20} )= \frac{629}{2400}= 0.26

P(A)= \frac{37}{120} = 0.308

P(B|A)= \frac{P(AnB)}{P(A)}= \frac{629/2400}{37/120} = \frac{17}{20} = 0.85

As you can see the probability of the event "The volunteer is female given that it was a student of the school class" means that you already know the selected volunteer was a student and only needed to calculate the probability of that student being female.

P(B)= 17/20

I hope this helps!

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Answer:

a

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b

    P( X \ge  6 )=  0.4516

c

   P( X <  4 )=  0.12694

Step-by-step explanation:

From the question we are told that

   The population proportion is   p = 0.53

    The sample size is  n = 10

Generally the distribution of the confidence of US adults in newspapers follows a binomial distribution  

i.e  

         X  \~ \ \ \  B(n , p)

and the probability distribution function for binomial  distribution is  

      P(X = x) =  ^{n}C_x *  p^x *  (1- p)^{n-x}

Here C stands for combination hence we are going to be making use of the combination function in our calculators        

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is exactly​ five is mathematically represented as

     P(X = 5) =  ^{10}C_5 *  0.53^5 *  (1- 0.53)^{10-5}

=>  P(X = 5) =  252 * 0.04182 *  0.023

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Generally the probability that the number of U.S. adults who have very little confidence in newspapers is  at least​ six is mathematically represented as

  P( X \ge  6 )=  P( X  = 6) +  P( X  = 7) +  P( X  = 8) +  P( X  = 9) +  P( X  = 10)

=> P( X \ge  6 )=  [^{10}C_6 *  [0.53]^6 *  (1- 0.53)^{10-6}] +  [^{10}C_7 *  [0.53]^7 *  (1- 0.53)^{10-7}] +  [^{10}C_8 *  [0.53]^8 *  (1- 0.53)^{10-8}] +  [^{10}C_9 *  [0.53]^9 *  (1- 0.53)^{10-9}] + [^{10}C_{10} *  [0.53]^{10} *  (1- 0.53)^{10-10}]

=> P( X \ge  6 )=  [0.227] +  [0.1464] +  [0.0619] +  [0.0155] + [0.00082]

=> P( X \ge  6 )=  0.4516

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is  less than four is mathematically represented as

    P( X <  4 )=  P( X  = 3) +  P( X  = 2) +  P( X  = 1) +  P( X  = 0)

=>  P( X <  4 )=  [^{10}C_3 *  0.53^3 *  (1- 0.53)^{10-3}] +  [^{10}C_2 *  0.53^2 *  (1- 0.53)^{10-2}] +  [^{10}C_1 *  0.53^1 *  (1- 0.53)^{10-1}] +  [^{10}C_0 *  0.53^0 *  (1- 0.53)^{10-0}]

=>  P( X <  4 )=  0.09051  +  0.03 +  0.0059 +  0.000526

=>  P( X <  4 )=  0.12694

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