Questions: Assume that random guesses are made for eight multiple choice questions on an SAT test, so that there are n=8 trials, each with probability of success (correct) given by p=0.45. Find the indicated probability for the number of correct answers. Find the probability that the number x of correct answers is fewer than 4 P(X<4)= (Round to four decimal places as needed.)

Solution Steps

We are tasked with finding the probability that the number \( X \) of correct answers on an SAT test, consisting of \( n = 8 \) multiple-choice questions with a probability of success \( p = 0.45 \), is fewer than 4. This can be expressed mathematically as \( P(X < 4) \).

To find \( P(X < 4) \), we need to calculate the probabilities for \( X = 0, 1, 2, \) and \( 3 \) using the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where \( q = 1 - p = 0.55 \).

Calculating each probability:

• For \( X = 0 \): \[ P(X = 0) = \binom{8}{0} \cdot (0.45)^0 \cdot (0.55)^8 = 0.0084 \]

• For \( X = 1 \): \[ P(X = 1) = \binom{8}{1} \cdot (0.45)^1 \cdot (0.55)^7 = 0.0548 \]

• For \( X = 2 \): \[ P(X = 2) = \binom{8}{2} \cdot (0.45)^2 \cdot (0.55)^6 = 0.1569 \]

• For \( X = 3 \): \[ P(X = 3) = \binom{8}{3} \cdot (0.45)^3 \cdot (0.55)^5 = 0.2568 \]

Now, we sum the probabilities calculated for \( X = 0, 1, 2, \) and \( 3 \):

\[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]

Substituting the values:

\[ P(X < 4) = 0.0084 + 0.0548 + 0.1569 + 0.2568 = 0.4769 \]

Final Answer