The Binomial Distribution1.The multiple choice section of a test consists of 5 questions.  Each question has 4 choices and is answered by guessing. a) What is the probability of answering the...

The Binomial Distribution

1.The multiple choice section of a test consists of 5 questions.  Each question has 4 choices and is answered by guessing.

a) What is the probability of answering the first 3 questions correctly an the last 2 questions incorrectly? Round your answer to the nearest one hundredth of a percent. 

b) What is the probability of answering any 3 questions correctly? Round your answer to the nearest one hundredth of a percent. 

c) Describe all the possible outcomes when the 5 questions are answered by guessing.  What is the probability of each of these outcomes? Round your answers to the nearest one tenth of a percent.

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embizze | High School Teacher | (Level 1) Educator Emeritus

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There are 5 questions with 4 possible answers for each.

(a) Determine the probability of getting the first three questions correct and missing the last two.

Using the counting principle we get `1/4*1/4*1/4*3/4*3/4=9/1024~~.88%`

(The first choice has a 1 in 4 chance of being correct as does the second and third. The last two each have a probability of 3 out of 4. The total probability is found by multiplying the probabilities.)

(b) Determine the probability of getting any three questions correct.

We use the binomial probability formula: `P(X)=(n!)/((n-X)!X!)*p^X*q^(n-X)` where P(X) is the probability of exactly X successes, n is the number of trials, p is the probability of success on a given trial and q is the probability of failure (or 1-p.) The formula can be written as `P(X)= ( _nC_x)p^x*q^(n-x)`

Then `P(3)=. _5C_3(.25)^3(.75)^2=10(1/64)(9/16)~~8.79%`

(c) The possibilities are

(1) All 5 correct: `P(5)= ._5C_5(.25)^5(.75)^0=1(1/1024)(1)=1/1024~~.10%`

(2) 4 correct : `P(4)= ._5C_4(.25)^4(.75)^1=5(1/256)(3/4)=15/1024~~1.46%`

(3) 3 correct:

calculated above as `P(3)~~8.79%`

(4) 2 correct:

`P(2)= ._5C_2(.25)^2(.75)^3=10(1/16)(27/64)=270/1024~~26.37%`

(5) 1 correct:

`P(1)= ._5C_1(.25)^1(.75)^4=5(1/4)(81)/(256)=405/1024~~39.55%`

(6) none correct:

`P(0)= ._5C_0(.25)^0(.75)^5=1(1)(243/1024)=243/1024~~23.73%`

As a quick check, the probabilities should sum to zero as we have included all possible outcomes.

.1+1.46+8.79+26.37+39.55+23.73=100%

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