stan guessed on all 10 of a multiple-choice quiz. each question has 4 answers. what is the probability that he got at least 2 questions correct?

Answer 1

Let's find probability of getting no correct answers or one correct answer.

Probability is ratio of event set size to sample set size.

Sample set size is 4¹⁰ (10 questions with 4 possible answers).

'Getting no correct answers' event size is 3¹⁰ (10 questions with 3 possible wrong answers).

'Getting one correct answer' event size is ₁₀C₁ * 3⁹ = 10* 3⁹ (there are 10 ways to select one correct answer and there are 3⁹ ways to select wrong answer on remaining 9 questions)).

P (of getting no correct answers or one correct answer) = (3¹⁰ + 10* 3⁹ ) / 4¹⁰ = (approx) = 0.244

So

P (of getting at least 2 correct answers ) = (approx) = 1- 0.244 = 0.756

Answer 2

This is a "choice" problem.

The probability of getting any question right is 1/4 = 0.25 (or 25%)--because each question has 4 options.

n = 10 (total # of questions)

k = 2 (2 correct)

p = 0.25 (probability correct)

q = 0.75 (probability incorrect, q=1-p)

(Note: * means "times")

so _nC_k * (p)^k *(q)^n-k

₁₀C₂ * (0.25)² * (0.75)⁸

where ₁₀C₂ = 10!/(2! * (10-2)!) = 10!/(2! * 8!)

=(10*9*8*7*6*5*....)/(2*1*8*7*6*5*4*3*2*1)

(cancel out like terms from top & bottom of fraction)

=5*9

=45

so it becomes

45 * (.25)² * (.75)⁸

Plug in your calculator

(if you have a graphing calculator you can do the ₁₀C₂ in your calculator)