Question

Assume that random guesses are made for nine multiple choice questions on an SAT​ test, so that there are nequals9 ​trials, each with probability of success​ (correct) given by pequals0.55. Find the indicated probability for the number of correct answers. Find the probability that the number x of correct answers is fewer than 4.

Answer 1

Answer:

$P(X=0)=(9C0)(0.55)^9 (1-0.55)^{9-0}=0.000757$

$P(X=1)=(9C1)(0.55)^9 (1-0.55)^{9-1}=0.0083$

$P(X=2)=(9C2)(0.55)^9 (1-0.55)^{9-2}=0.0407$

$P(X=3)=(9C3)(0.55)^9 (1-0.55)^{9-3}=0.1160$

And adding we got:

$P(X < 4) = 0.000757 +0.0083+0.0407 +0.1160= 0.2626$

Step-by-step explanation:

Let X the random variable of interest "number of correct answers", on this case we now that:

$X \sim Binom(n=9, p=0.55)$

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:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

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

And we can find the individual probabilities:

$P(X=0)=(9C0)(0.55)^9 (1-0.55)^{9-0}=0.000757$

$P(X=1)=(9C1)(0.55)^9 (1-0.55)^{9-1}=0.0083$

$P(X=2)=(9C2)(0.55)^9 (1-0.55)^{9-2}=0.0407$

$P(X=3)=(9C3)(0.55)^9 (1-0.55)^{9-3}=0.1160$

And adding we got:

$P(X < 4) = 0.000757 +0.0083+0.0407 +0.1160= 0.2626$