Mimi plans to make a random guess at a test containing 5 true-or-false questions. Mimi plans to...

Question:

Mimi plans to make a random guess at a test containing 5 true-or-false questions. Mimi plans to make a random guess at 5 true-or-false questions.

What is the probability that Mimi will pass the test (i.e. get at least 3 correct answers) by random guess?

Binomial Probability:

Binomial probability in statistics is the probability of exactly x successes on n repeated trials in an experiment that has two possible outcomes (success and failure).

Let say we have a discrete random variable {eq}x {/eq} with the probability of success in any trial is given as {eq}p {/eq}, then the probability of having {eq}x {/eq} number in the {eq}n {/eq} trials is given as:

{eq}P(X=x) = _nC_x \cdot p^{x} (1-p)^{n-x} {/eq}, where {eq}_nC_x = {n \choose x} = \dfrac{n!}{(n-x)!\cdot x!} {/eq}

Let the random variable {eq}x {/eq} be the number of correct answers Mimi gets. So {eq}x = 0,1,2,3,4,\text{ or } 5. {/eq} The number of trials {eq}n = 5, P(\text{ True })=p =\dfrac{1}{2}=0.5 {/eq}

What is the probability that Mimi will pass the test (i.e. get at least 3 correct answers) by random guess?

At leats 3 means greater than or equal to 3 ({eq}x =3,4,\text{ or } 5 {/eq}). So we can write the probability mathematically as:

{eq}P(x \geq 3)= P(x=3) + P(x=4) + P(x=5) \\ P(X=x) = _nC_x \cdot p^{x} (1-p)^{n-x} \\ P(x=3) = _5C_3 \cdot 0.5^{3} (1-0.5)^{5-3} = \dfrac{5!}{(5-3)!\times 3!}\times 0.5^{3} (1-0.5)^{2} = 0.3125 \\ P(x=4) = _5C_4 \cdot 0.5^{4} (1-0.5)^{5-4} = \dfrac{5!}{(5-4)!\times 4!}\times 0.5^{4} (1-0.5)^{1} = 0.15625\\ P(x=5) = _5C_5 \cdot 0.5^{5} (1-0.5)^{5-5} = \dfrac{5!}{(5-5)!\times 5!}\times 0.5^{5} (1-0.5)^{0} = 0.03125\\ P(x \geq 3)= P(x=3) + P(x=4) + P(x=5) = 0.3125 + 0.15625 + 0.03125 = 0.5\\ {/eq}

Therefore, the probability that Mimi will pass the test (i.e. get at least 3 correct answers) by random guess is {eq}0.5 {/eq} or 50%.