Respuesta :
The probability of getting first 4 right and the rest wrong = (1/4)^4*(3/4)^16
there are 20C5 ways = 4845 ways of getting 4 right out of 20 so required propability is (1/4)^4 * (3/4)^16 * 4845 = 0.190 answer
there are 20C5 ways = 4845 ways of getting 4 right out of 20 so required propability is (1/4)^4 * (3/4)^16 * 4845 = 0.190 answer
Answer:
There is a 18.97% probability that the student gets exactly four correct.
Step-by-step explanation:
For each question, there are only two possible outcomes. Either the student gets it correct, or he gets it wrong. This means that we use the binomial probability distribution to solve this problem.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
20 questions, so [tex]n = 20[/tex]
Each question has four options and one correct answer, that is guessed, so [tex]p = \frac{1}{4} = 0.25[/tex]
Find the probability that the student gets exactly four correct.
This is P(X = 4).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{20,4}.(0.25)^{4}.(0.75)^{16} = 0.1897[/tex]
There is a 18.97% probability that the student gets exactly four correct.
