Question

a test consists of 10 true/false questions. to pass the test a student must answer at least 6 questions correctly. if a student guesses on each question, what is the probability that the student will pass the test? round to three decimal places

a. 0.205
b. 0.172
c. 0.377
d. 0.828

Explanation:

Step1: Identify probability of single - question

Each question is a true - false question, so the probability of getting a single question correct by guessing is $p=\frac{1}{2}=0.5$, and the probability of getting it wrong is $q = 1 - p=0.5$. The number of questions $n = 10$. We use the binomial probability formula $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Calculate probabilities for $k = 6,7,8,9,10$

For $k = 6$:
$C(10,6)=\frac{10!}{6!(10 - 6)!}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210$
$P(X = 6)=C(10,6)\times(0.5)^{6}\times(0.5)^{4}=210\times(0.5)^{10}$
For $k = 7$:
$C(10,7)=\frac{10!}{7!(10 - 7)!}=\frac{10\times9\times8}{3\times2\times1}=120$
$P(X = 7)=C(10,7)\times(0.5)^{7}\times(0.5)^{3}=120\times(0.5)^{10}$
For $k = 8$:
$C(10,8)=\frac{10!}{8!(10 - 8)!}=\frac{10\times9}{2\times1}=45$
$P(X = 8)=C(10,8)\times(0.5)^{8}\times(0.5)^{2}=45\times(0.5)^{10}$
For $k = 9$:
$C(10,9)=\frac{10!}{9!(10 - 9)!}=10$
$P(X = 9)=C(10,9)\times(0.5)^{9}\times(0.5)^{1}=10\times(0.5)^{10}$
For $k = 10$:
$C(10,10)=\frac{10!}{10!(10 - 10)!}=1$
$P(X = 10)=C(10,10)\times(0.5)^{10}\times(0.5)^{0}=1\times(0.5)^{10}$

Step3: Sum up the probabilities

$P(X\geq6)=P(X = 6)+P(X = 7)+P(X = 8)+P(X = 9)+P(X = 10)$
$=(210 + 120+45 + 10+1)\times(0.5)^{10}$
$=386\times(0.5)^{10}$
$=386\times\frac{1}{1024}\approx0.377$

Answer:

C. 0.377