Question

grade p(grade) a 0.1 b 0.3 c 0.35 d 0.05 f? 2a. if i randomly choose 1 student at random from the distribution, what is the probability he/she got an f? round to the tenth. 0.2 2b. if i randomly choose 1 student at random from the distribution, what is the probability he/she did not get an a or b? round to the tenth. 0.6 2c. if i randomly choose 1 student at random from the distribution, what is the probability he/she got an a and a b?

Explanation:

Step1: Use the property of probability distribution

The sum of all probabilities in a probability distribution is \(1\). Let \(P(F)\) be the probability of getting an \(F\). Then \(P(A)+P(B)+P(C)+P(D)+P(F) = 1\).

Step2: Substitute the known values

We know \(P(A)=0.1\), \(P(B) = 0.3\), \(P(C)=0.35\), \(P(D)=0.05\). Substituting into the equation: \(0.1+0.3 + 0.35+0.05+P(F)=1\).

Step3: Simplify the left - hand side

\(0.1+0.3+0.35 + 0.05=0.8\). So the equation becomes \(0.8+P(F)=1\).

Step4: Solve for \(P(F)\)

Subtract \(0.8\) from both sides: \(P(F)=1 - 0.8=0.2\).

For part \(2b\):

Step1: Use the complement rule

The probability of not getting an \(A\) or \(B\) is \(P(\text{not }A\text{ or }B)=1 - P(A)-P(B)\).

Step2: Substitute the values of \(P(A)\) and \(P(B)\)

\(P(A) = 0.1\) and \(P(B)=0.3\). So \(P(\text{not }A\text{ or }B)=1-(0.1 + 0.3)=1 - 0.4=0.6\).

For part \(2c\):

Step1: Use the addition rule for mutually exclusive events

Since a student cannot get an \(A\) and a \(B\) simultaneously (mutually exclusive events), \(P(A\cap B)=0\).

Answer:

2a. \(0.2\)
2b. \(0.6\)
2c. \(0\)