Please help!!A multiple-choice test consists of 28 questions with possible answers of a,b,c,d. Estimate the probability that with random guessing, the number of correct answers is at least 12.

Answer:Approximately 0.0294.Step-by-step explanation:Assume that there's only one correct choice in each question. The chance of getting a question correct by random guess is 1/4.The chance of getting a question wrong by random guess is 3/4.What's the probability that exactly 12 answers are correct?12 out of the 28 answers need to be correct. [tex]\displaystyle \left(\frac{1}{4}\right)^{12}[/tex].The other 28 - 12 answers need to be incorrect. Multiply by [tex]\displaystyle \left(\frac{3}{4}\right)^{28 - 12}[/tex].There are more than one way of choosing 12 answers out of 28 without an order. Multiply by the combination "12-choose-28" [tex]\displaystyle \left(\begin{array}{c}12\\28\end{array}\right)[/tex].The probability of getting exactly 12 answers correct is:[tex]\displaystyle \left(\frac{1}{4}\right)^{12} \times \left(\frac{3}{4}\right)^{28 - 12}\times \left(\begin{array}{c}28\\12\end{array}\right)\approx 0.0182[/tex].With the same logic, the probability of getting [tex]x[/tex] ([tex]x\in \mathbb{Z}[/tex], [tex]12\le x\le 28[/tex]) correct out of the 28 random answers will be[tex]\displaystyle \left(\frac{1}{4}\right)^{x} \times \left(\frac{3}{4}\right)^{28 - x}\times \left(\begin{array}{c}28\\x\end{array}\right)[/tex].The probability of getting at least 12 correct out of 28 random answers is the sum of the probability of getting exactly 12 correct out of 28, plusthe probability of getting exactly 13 correct out of 28, plusthe probability of getting exactly 14 correct out of 28, plusthe probability of getting exactly 15 correct out of 28, plusthe probability of getting exactly 16 correct out of 28, plus... all the way to the probability of getting exactly 28 correct out of 28.The Sigma notation might help:[tex]\displaystyle \sum_{x = 12}^{28}{\left[\left(\frac{1}{4}\right)^{x} \times \left(\frac{3}{4}\right)^{28 - x}\times \left(\begin{array}{c}x\\28\end{array}\right)\right]}[/tex].Evaluate this sum (preferably with a calculator)[tex]\displaystyle \sum_{x = 12}^{28}{\left[\left(\frac{1}{4}\right)^{x} \times \left(\frac{3}{4}\right)^{28 - x}\times \left(\begin{array}{c}x\\28\end{array}\right)\right]} \approx 0.0294[/tex].