Question

Assume that random guesses are made for seven multiple choice questions on an SAT test, so that there are n=7 trials, each with probability of success (correct) given by p055 Find the indicated probability for the number of correct answers
Find the probability that the number of correct answers is fewer than 4
PX<4)=0
(Round to four decimal places as needed)

Answer 1

Answer:

P(X < 4) = 0.5

Step-by-step explanation:

For each question, there are only two possible outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent of any other question. This means that we use the binomial probability distribution to solve this question.

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.

In this question, we have that:

$n = 7, p = 0.5$

Find the probability that the number of correct answers is fewer than 4:

This is

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

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

$P(X = 0) = C_{7,0}.(0.5)^{0}.(0.5)^{7} = 0.0078$

$P(X = 1) = C_{7,1}.(0.5)^{1}.(0.5)^{6} = 0.0547$

$P(X = 2) = C_{7,2}.(0.5)^{2}.(0.5)^{5} = 0.1641$

$P(X = 3) = C_{7,3}.(0.5)^{3}.(0.5)^{4} = 0.2734$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0078 + 0.0547 + 0.1641 + 0.2734 = 0.5$

So

P(X < 4) = 0.5