A multiple-choice test contains 25 questions, each with four answers. Assume that a student just guesses on each question. (a) What is the probability that the student answers more than 20 questions correctly? (b) What is the probability that the student answers fewer than 5 questions correctly?

Respuesta :

Answer:

(a) The probability of guessing 20 answers correctly is 1.15*10^-8 or 0.000001%

(b) The probability of answer fewer than 5 questions correctly is 0.2138 or 21.38%.

Step-by-step explanation:

If the student is guessing, he or she has only 1 out of 4 chances of answer correctly. This is equivalent to a probability of success p=0.25.

For this problem we can use the binomial distribution to calculate the probabilities.

(a) What is the probability that the student answers more than 20 questions correctly?

We have a total of 25 questions. The probability of guessing 20 answers correctly is

[tex]P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\\\P(20)=\frac{25!}{20!5!}*0.25^{20}*0.75^5\\\\P(20)=53130* 9.095*10^{-3}*0.2373\\\\P(20)=1.15*10^{-8}[/tex]

The probability of guessing 20 answers correctly is 1.15*10^-8 or 0.000001% .

(b) The probability of answer fewer than 5 questions correctly is the sum of:

1) Answering 0 questions correctly

2) Answering 1 questions correctly

3) Answering 2 questions correctly

4) Answering 3 questions correctly

5) Answering 4 questions correctly

This probabilities can be calculated as

[tex]P(0)=\frac{25!}{0!25!}*0.25^0*0.75^{25}=1*1*7.53*10^{-4}=0.0008\\\\P(1)=\frac{25!}{1!24!}*0.25^1*0.75^{24}= 0.0063  

\\\\P(2)=\frac{25!}{2!23!}*0.25^2*0.75^{23}= 0.0251  

\\\\P(3)=\frac{25!}{3!22!}*0.25^3*0.75^{22}= 0.0641  

\\\\P(4)=\frac{25!}{4!21!}*0.25^4*0.75^{21}= 0.1175  

[/tex]

Then the sum of the probabilty of this events is

[tex]P(x<5)=P(0)+P(1)+P(2)+P(3)+P(4)[/tex]

[tex]P(x<5)=0.0008+0.0063+0.0251+0.0641+0.1175=0.2138[/tex]

The probability of answer fewer than 5 questions correctly is 0.2138 or 21.38%.