Question

cards example marbles table
total: blue red green
22 1 6 15 glass
27 11 13 3 steel
32 12 20 0 plastic
81 24 39 18 total
1 fill in the blank 3 points determine p(glass). (enter as fraction.) 5/6
find p(glass | green). 22/81
did the condition of being green increase/decrease the likelihood of drawing a glass marble? doesnt affect
2 fill in the blank 3 points determine p(red). (enter as fraction.)
find p(red | steel).
did the condition of being green increase/decrease the likelihood of drawing a glass marble? choose your answer...
3 multiple choice 1 point

Explanation:

Step1: Recall probability formula

The formula for probability is $P(A)=\frac{n(A)}{n(S)}$, where $n(A)$ is the number of elements in event $A$ and $n(S)$ is the total number of elements in the sample - space. For conditional probability, $P(A|B)=\frac{P(A\cap B)}{P(B)}$.

Step2: Calculate $P(\text{red})$

The total number of marbles $n(S) = 81$. The number of red marbles $n(\text{red})=39$. So $P(\text{red})=\frac{39}{81}=\frac{13}{27}$.

Step3: Calculate $P(\text{red}|\text{steel})$

The number of steel marbles $n(\text{steel}) = 27$. The number of red - steel marbles $n(\text{red}\cap\text{steel}) = 13$. By the formula for conditional probability $P(\text{red}|\text{steel})=\frac{n(\text{red}\cap\text{steel})}{n(\text{steel})}=\frac{13}{27}$.

Step4: Analyze the relationship between being green and being glass

The number of green marbles $n(\text{green}) = 18$, the number of green - glass marbles $n(\text{green}\cap\text{glass}) = 15$, the number of glass marbles $n(\text{glass}) = 22$.
$P(\text{glass})=\frac{22}{81}$, $P(\text{glass}|\text{green})=\frac{n(\text{green}\cap\text{glass})}{n(\text{green})}=\frac{15}{18}=\frac{5}{6}$. Since $\frac{5}{6}\approx0.833$ and $\frac{22}{81}\approx0.272$, the condition of being green increases the likelihood of drawing a glass marble.

Answer:

1. $\frac{22}{81}$
2. $\frac{39}{81}$ (or $\frac{13}{27}$), $\frac{13}{27}$, increases
3. N/A (no multiple - choice options provided to choose from for the last part of question 3)