Question

1. at laras school, 75% of the student body is male and 35% of the students walk to school. assume that these two events are independent. if a student from laras school is selected at random, determine the following probabilities.

a. create a probability area model or tree diagram to represent this situation.
b. p(student is female)
c. p(student is male and does not walk to school)
d. p(student is female or does not walk to school)
e. identify the sample space in parts (c) and (d) above as a union or an intersection.

Explanation:

Step1: Define probabilities

Let $P(M) = 0.75$ (probability of male), so the probability of female $P(F)=1 - P(M)=1 - 0.75 = 0.25$. Let $P(W)=0.35$ (probability of walking to school), then $P(\overline{W})=1 - P(W)=1 - 0.35 = 0.65$. Since the two - events are independent, for two independent events $A$ and $B$, $P(A\cap B)=P(A)\times P(B)$.

Step2: Probability of female

$P(\text{student is female})=1 - P(\text{student is male})$. Since $P(\text{student is male}) = 0.75$, then $P(\text{student is female})=0.25$.

Step3: Probability of male and not walking

$P(\text{male and not walking})=P(M)\times P(\overline{W})$. Substitute $P(M) = 0.75$ and $P(\overline{W}) = 0.65$ into the formula, we get $P(M)\times P(\overline{W})=0.75\times0.65 = 0.4875$.

Step4: Probability of female or not walking

Use the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Here $A$ is the event of being female and $B$ is the event of not walking. $P(F\cup\overline{W})=P(F)+P(\overline{W})-P(F)\times P(\overline{W})$. Substitute $P(F) = 0.25$ and $P(\overline{W}) = 0.65$ into the formula: $P(F\cup\overline{W})=0.25 + 0.65-0.25\times0.65=0.25 + 0.65 - 0.1625=0.7375$.

Step5: Identify sample - space type

In part (c), $P(\text{male and not walking})$ is an intersection of two events, so it is an intersection. In part (d), $P(\text{female or not walking})$ is a union of two events, so it is a union.

Answer:

b. $0.25$
c. $0.4875$
d. $0.7375$
e. c: intersection; d: union