there are 2 red, 2 orange, 5 yellow, 3 green marbles in a bag. an experiment was conducted 72 times where a…

there are 2 red, 2 orange, 5 yellow, 3 green marbles in a bag. an experiment was conducted 72 times where a marble was drawn randomly from the bag, twice, with replacement. the results are shown in the table.\n| outcome | number of times | outcome | number of times |\n| ---- | ---- | ---- | ---- |\n| red, red | 3 | yellow, red | 8 |\n| red, orange | 1 | yellow, orange | 5 |\n| red, yellow | 6 | yellow, yellow | 14 |\n| red, green | 5 | yellow, green | 7 |\n| orange, red | 0 | green, red | 3 |\n| orange, orange | 3 | green, orange | 3 |\n| orange, yellow | 5 | green, yellow | 5 |\n| orange, green | 2 | green, green | 2 |\nselect all of the outcomes of the experiment in which the experimental probability matches the theoretical probability.\n□ p(red, yellow)\n□ p(red, green)\n□ p(orange, red)\n□ p(orange, yellow)\n□ p(yellow, orange)\n□ p(yellow, green)\n□ p(green, red)\n□ p(green, orange)

there are 2 red, 2 orange, 5 yellow, 3 green marbles in a bag. an experiment was conducted 72 times where a marble was drawn randomly from the bag, twice, with replacement. the results are shown in the table.\n| outcome | number of times | outcome | number of times |\n| ---- | ---- | ---- | ---- |\n| red, red | 3 | yellow, red | 8 |\n| red, orange | 1 | yellow, orange | 5 |\n| red, yellow | 6 | yellow, yellow | 14 |\n| red, green | 5 | yellow, green | 7 |\n| orange, red | 0 | green, red | 3 |\n| orange, orange | 3 | green, orange | 3 |\n| orange, yellow | 5 | green, yellow | 5 |\n| orange, green | 2 | green, green | 2 |\nselect all of the outcomes of the experiment in which the experimental probability matches the theoretical probability.\n□ p(red, yellow)\n□ p(red, green)\n□ p(orange, red)\n□ p(orange, yellow)\n□ p(yellow, orange)\n□ p(yellow, green)\n□ p(green, red)\n□ p(green, orange)

Answer

Explanation:

Step1: Calculate total number of marbles

$2 + 2+5 + 3=12$

Step2: Calculate theoretical probabilities for each compound - event

For two - draw events with replacement, if event A and B are the first and second draws respectively, $P(A,B)=P(A)\times P(B)$. For example, $P(\text{Red},\text{Yellow})=\frac{2}{12}\times\frac{5}{12}=\frac{10}{144}=\frac{5}{72}$ $P(\text{Red},\text{Green})=\frac{2}{12}\times\frac{3}{12}=\frac{6}{144}=\frac{1}{24}$ $P(\text{Orange},\text{Red})=\frac{2}{12}\times\frac{2}{12}=\frac{4}{144}=\frac{1}{36}$ $P(\text{Orange},\text{Yellow})=\frac{2}{12}\times\frac{5}{12}=\frac{10}{144}=\frac{5}{72}$ $P(\text{Yellow},\text{Orange})=\frac{5}{12}\times\frac{2}{12}=\frac{10}{144}=\frac{5}{72}$ $P(\text{Yellow},\text{Green})=\frac{5}{12}\times\frac{3}{12}=\frac{15}{144}=\frac{5}{48}$ $P(\text{Green},\text{Red})=\frac{3}{12}\times\frac{2}{12}=\frac{6}{144}=\frac{1}{24}$ $P(\text{Green},\text{Orange})=\frac{3}{12}\times\frac{2}{12}=\frac{6}{144}=\frac{1}{24}$

Step3: Calculate experimental probabilities

Experimental probability of an event $E$ is $P_{exp}(E)=\frac{\text{Number of times }E\text{ occurred}}{\text{Total number of trials}}$. Total number of trials = 72. $P_{exp}(\text{Red},\text{Yellow})=\frac{6}{72}=\frac{1}{12}\neq\frac{5}{72}$ $P_{exp}(\text{Red},\text{Green})=\frac{5}{72}\neq\frac{1}{24}$ $P_{exp}(\text{Orange},\text{Red})=\frac{0}{72}=0\neq\frac{1}{36}$ $P_{exp}(\text{Orange},\text{Yellow})=\frac{5}{72}$ $P_{exp}(\text{Yellow},\text{Orange})=\frac{5}{72}$ $P_{exp}(\text{Yellow},\text{Green})=\frac{7}{72}\neq\frac{5}{48}$ $P_{exp}(\text{Green},\text{Red})=\frac{3}{72}=\frac{1}{24}$ $P_{exp}(\text{Green},\text{Orange})=\frac{3}{72}=\frac{1}{24}$

Answer:

P(Orange, Yellow), P(Yellow, Orange), P(Green, Red), P(Green, Orange)