probability and independent / dependent events.\nor probability\np(a ∪ b)\np(a or b)=p(a)+p(b)-p(a and…

probability and independent / dependent events.\nor probability\np(a ∪ b)\np(a or b)=p(a)+p(b)-p(a and b)\n1. p(a) = 0.25\np(b) = 0.20\np(a and b) = 0.40\nwhat is p(a or b)?\nand probability\np(a ∩ b)\np(a and b)=p(a)·p(b)\n2. at a school, being in 9th grade and being in 12th grade are mutually exclusive events. choosing a student at random, the probability of choosing in 9th grade is 0.26. the probability that a randomly selected student is in 12th grade is 0.22.\nwhat is the probability that a randomly selected student is in 9th or 12th grade?

probability and independent / dependent events.\nor probability\np(a ∪ b)\np(a or b)=p(a)+p(b)-p(a and b)\n1. p(a) = 0.25\np(b) = 0.20\np(a and b) = 0.40\nwhat is p(a or b)?\nand probability\np(a ∩ b)\np(a and b)=p(a)·p(b)\n2. at a school, being in 9th grade and being in 12th grade are mutually exclusive events. choosing a student at random, the probability of choosing in 9th grade is 0.26. the probability that a randomly selected student is in 12th grade is 0.22.\nwhat is the probability that a randomly selected student is in 9th or 12th grade?

Answer

Explanation:

Step1: Recall the formula for P(A or B)

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Step2: Solve for question 1

Given $P(A) = 0.25$, $P(B)=0.20$ and $P(A\cap B)=0.40$. Substitute into the formula: $P(A\cup B)=0.25 + 0.20-0.40$ $P(A\cup B)=0.05$

Step3: Analyze question 2

Since being in 9th - grade and 12th - grade are mutually - exclusive events, $P(A\cap B) = 0$. Let $A$ be the event of being in 9th grade and $B$ be the event of being in 12th grade. Given $P(A)=0.26$ and $P(B)=0.22$.

Step4: Solve for question 2

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, substitute $P(A\cap B) = 0$: $P(A\cup B)=0.26+0.22 - 0$ $P(A\cup B)=0.48$

Answer:

  1. $P(A\cup B)=0.05$
  2. $P(A\cup B)=0.48$