Suppose a student takes a $10$-question true/false exam and also guesses in ~ every question. What is the probability that... A) the student answers every concern incorrectly? b) the students answers at the very least one question correctly? c) the college student answers precisely 5 questions correctly?

So far I have solved a. I acquired $(1/2)^10$.For b I"m not certain what come do.For c would certainly the answer it is in $(1/2)^5$?

a) Is fine.

You are watching: A student takes a 10 question true-false exam and guesses on each question

b) an alert that the complementary event of $A =$"at least one right" is $\bar A =$"all wrong". Hence$$P(A) = 1-P(\bar A) = 1-\left(\frac12\right)^10.$$

c) No. You have to count all the spots wherein you gained them right. Notice that there space $\binom105$ methods to select the concerns that room correct. We additionally know the $5$ room wrong (with possibility $1/2$ every time) and $5$ are ideal (with chance $1/2$ every time). Hence$$P(\textExactly 5 right) = \binom105\left(\frac12\right)^5\left(\frac12\right)^5.$$

The number of correct answer in $10$ tries adheres to a binomial distribution.

Given that the probability of comment a question correctly is $\dfrac12$:

a) the college student answers every inquiry incorrectly?

$$\left(1-\frac12\right)^10$$

b) the students answer at least one concern correctly?

$$1-\left(1-\frac12\right)^10$$

c) the student answers exactly 5 concerns correctly?

$$\binom105\cdot\left(\frac12\right)^5\cdot\left(1-\frac12\right)^10-5$$

Let $X$ be a random variable the maps the variety of correct answers guessed come its probability. $X$ is discrete and thus $X \sim Bionomial(n=10, p = \frac12)$. Simply plug in the corresponding $P(X = a)$ and calculate via. The binomial circulation equation come compute your wanted answer. For component b that is $P(X \geq 1) = 1 - P(X =0)$

For b it will be $$\sum\frac10\choose i2^10$$ where $ns \in <1,10>$ as its atleast 1 exactly . Because that c it will certainly be any kind of $5$ native $10$ therefore it will certainly be $$\frac10\choose 52^10$$

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