Question

Assume that random guesses are made for seven multiple choice questions on an SAT​ test, so that there are nequals7 ​trials, each with probability of success​ (correct) given by pequals0.25. Find the indicated probability for the number of correct answers.

Answer 1

`P(X < 4)=0.1335+0.3115+0.3115+0.1730=0.9295`

Explanation:

Question assumed: Find the probability that the number x of correct answers is fewer than 4.

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=7, p=0.25)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

And we want to find this probability:

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

`P(X=0)=(7C0)(0.25)^0 (1-0.25)^(7-0)=0.1335`

`P(X=1)=(7C1)(0.25)^1 (1-0.25)^(7-1)=0.3115`

`P(X=2)=(7C2)(0.25)^2 (1-0.25)^(7-2)=0.3115`

`P(X=3)=(7C3)(0.25)^3 (1-0.25)^(7-3)=0.1730`

`P(X < 4)=0.1335+0.3115+0.3115+0.1730=0.9295`