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?
b) the students answer at least one concern correctly?
c) the student answers exactly 5 concerns correctly?
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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