Question

1. (4 points) a college professor hands out a list of 10 questions, five of which will appear on the final exam for the course (the professor chooses the five questions at random from the 10). one of the students taking the course is pressed for time, and can prepare for only seven of the 10 questions. what is the probability that the student will be prepared for all five questions that appear on the final exam? 9) (2 points) suppose a and b are two events for which p(a) =.4, p(b) =.6, and p(a∪b) =.76. (a) are a and b incompatible? explain the reason. (b) are a and b independent? explain the reason.

Explanation:

Step1: Calculate total number of ways to choose 5 questions out of 10

The number of combinations of choosing $r$ items from $n$ items is given by the formula $C(n,r)=\frac{n!}{r!(n - r)!}$. Here, $n = 10$ and $r=5$, so the total number of ways to choose 5 questions out of 10 is $C(10,5)=\frac{10!}{5!(10 - 5)!}=\frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1}=252$.

Step2: Calculate number of favorable ways

The student prepares for 7 questions. The number of ways to choose 5 questions from the 7 - prepared questions is $C(7,5)=\frac{7!}{5!(7 - 5)!}=\frac{7\times6}{2\times1}=21$.

Step3: Calculate the probability

The probability $P$ that the student is prepared for all 5 questions on the exam is the number of favorable ways divided by the total number of ways. So $P=\frac{C(7,5)}{C(10,5)}=\frac{21}{252}=\frac{1}{12}$.

for 9(a):
Recall the formula for the probability of the union of two events: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. If $A$ and $B$ are incompatible (mutually - exclusive), then $P(A\cap B) = 0$. Given $P(A)=0.4$, $P(B)=0.6$ and $P(A\cup B)=0.76$. We can find $P(A\cap B)$ using the formula $P(A\cap B)=P(A)+P(B)-P(A\cup B)$. Substituting the values, we get $P(A\cap B)=0.4 + 0.6-0.76=0.24
eq0$. So $A$ and $B$ are not incompatible.

for 9(b):
Two events $A$ and $B$ are independent if $P(A\cap B)=P(A)\times P(B)$. We know $P(A)=0.4$, $P(B)=0.6$, so $P(A)\times P(B)=0.4\times0.6 = 0.24$. And from part (a), we found $P(A\cap B)=0.24$. Since $P(A\cap B)=P(A)\times P(B)$, events $A$ and $B$ are independent.

Answer:

$\frac{1}{12}$